Probability density function

In probability theory, a probability density function (PDF), density function, or density of an absolutely continuous random variable, is a function whose value at any given sample (or point) in the sample space (the set of possible values taken by the random variable) can be interpreted as providing a relative likelihood that the value of the random variable would be equal to that sample.^[2]^[3] Probability density is the probability per unit length, in other words, while the absolute likelihood for a continuous random variable to take on any particular value is 0 (since there is an infinite set of possible values to begin with), the value of the PDF at two different samples can be used to infer, in any particular draw of the random variable, how much more likely it is that the random variable would be close to one sample compared to the other sample.

In a more precise sense, the PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value. This probability is given by the integral of this variable's PDF over that range—that is, it is given by the area under the density function but above the horizontal axis and between the lowest and greatest values of the range. The probability density function is nonnegative everywhere, and the area under the entire curve is equal to 1.

The terms probability distribution function and probability function have also sometimes been used to denote the probability density function. However, this use is not standard among probabilists and statisticians. In other sources, "probability distribution function" may be used when the probability distribution is defined as a function over general sets of values or it may refer to the cumulative distribution function, or it may be a probability mass function (PMF) rather than the density. "Density function" itself is also used for the probability mass function, leading to further confusion.^[4] In general though, the PMF is used in the context of discrete random variables (random variables that take values on a countable set), while the PDF is used in the context of continuous random variables.

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In the last video, I introduced you to the notion of-- well, really we started with the random variable. And then we moved on to the two types of random variables. You had discrete, that took on a finite number of values. And the these, I was going to say that they tend to be integers, but they don't always have to be integers. You have discrete, so finite meaning you can't have an infinite number of values for a discrete random variable. And then we have the continuous, which can take on an infinite number. And the example I gave for continuous is, let's say random variable x. And people do tend to use-- let me change it a little bit, just so you can see it can be something other than an x. Let's have the random variable capital Y. They do tend to be capital letters. Is equal to the exact amount of rain tomorrow. And I say rain because I'm in northern California. It's actually raining quite hard right now. We're short right now, so that's a positive. We've been having a drought, so that's a good thing. But the exact amount of rain tomorrow. And let's say I don't know what the actual probability distribution function for this is, but I'll draw one and then we'll interpret it. Just so you can kind of think about how you can think about continuous random variables. So let me draw a probability distribution, or they call it its probability density function. And we draw like this. And let's say that there is-- it looks something like this. Like that. All right, and then I don't know what this height is. So the x-axis here is the amount of rain. Where this is 0 inches, this is 1 inch, this is 2 inches, this is 3 inches, 4 inches. And then this is some height. Let's say it peaks out here at, I don't know, let's say this 0.5. So the way to think about it, if you were to look at this and I were to ask you, what is the probability that Y-- because that's our random variable-- that Y is exactly equal to 2 inches? That Y is exactly equal to two inches. What's the probability of that happening? Well, based on how we thought about the probability distribution functions for the discrete random variable, you'd say OK, let's see. 2 inches, that's the case we care about right now. Let me go up here. You'd say it looks like it's about 0.5. And you'd say, I don't know, is it a 0.5 chance? And I would say no, it is not a 0.5 chance. And before we even think about how we would interpret it visually, let's just think about it logically. What is the probability that tomorrow we have exactly 2 inches of rain? Not 2.01 inches of rain, not 1.99 inches of rain. Not 1.99999 inches of rain, not 2.000001 inches of rain. Exactly 2 inches of rain. I mean, there's not a single extra atom, water molecule above the 2 inch mark. And not as single water molecule below the 2 inch mark. It's essentially 0, right? It might not be obvious to you, because you've probably heard, oh, we had 2 inches of rain last night. But think about it, exactly 2 inches, right? Normally if it's 2.01 people will say that's 2. But we're saying no, this does not count. It can't be 2 inches. We want exactly 2. 1.99 does not count. Normally our measurements, we don't even have tools that can tell us whether it is exactly 2 inches. No ruler you can even say is exactly 2 inches long. At some point, just the way we manufacture things, there's going to be an extra atom on it here or there. So the odds of actually anything being exactly a certain measurement to the exact infinite decimal point is actually 0. The way you would think about a continuous random variable, you could say what is the probability that Y is almost 2? So if we said that the absolute value of Y minus is 2 is less than some tolerance? Is less than 0.1. And if that doesn't make sense to you, this is essentially just saying what is the probability that Y is greater than 1.9 and less than 2.1? These two statements are equivalent. I'll let you think about it a little bit. But now this starts to make a little bit of sense. Now we have an interval here. So we want all Y's between 1.9 and 2.1. So we are now talking about this whole area. And area is key. So if you want to know the probability of this occurring, you actually want the area under this curve from this point to this point. And for those of you who have studied your calculus, that would essentially be the definite integral of this probability density function from this point to this point. So from-- let me see, I've run out of space down here. So let's say if this graph-- let me draw it in a different color. If this line was defined by, I'll call it f of x. I could call it p of x or something. The probability of this happening would be equal to the integral, for those of you who've studied calculus, from 1.9 to 2.1 of f of x dx. Assuming this is the x-axis. So it's a very important thing to realize. Because when a random variable can take on an infinite number of values, or it can take on any value between an interval, to get an exact value, to get exactly 1.999, the probability is actually 0. It's like asking you what is the area under a curve on just this line. Or even more specifically, it's like asking you what's the area of a line? An area of a line, if you were to just draw a line, you'd say well, area is height times base. Well the height has some dimension, but the base, what's the width the a line? As far as the way we've defined a line, a line has no with, and therefore no area. And it should make intuitive sense. That the probability of a very super-exact thing happening is pretty much 0. That you really have to say, OK what's the probably that we'll get close to 2? And then you can define an area. And if you said oh, what's the probability that we get someplace between 1 and 3 inches of rain, then of course the probability is much higher. The probability is much higher. It would be all of this kind of stuff. You could also say what's the probability we have less than 0.1 of rain? Then you would go here and if this was 0.1, you would calculate this area. And you could say what's the probability that we have more than 4 inches of rain tomorrow? Then you would start here and you'd calculate the area in the curve all the way to infinity, if the curve has area all the way to infinity. And hopefully that's not an infinite number, right? Then your probability won't make any sense. But hopefully if you take this sum it comes to some number. And we'll say there's only a 10% chance that you have more than 4 inches tomorrow. And all of this should immediately lead to one light bulb in your head, is that the probability of all of the events that might occur can't be more than 100%. Right? All the events combined-- there's a probability of 1 that one of these events will occur. So essentially, the whole area under this curve has to be equal to 1. So if we took the integral of f of x from 0 to infinity, this thing, at least as I've drawn it, dx should be equal to 1. For those of you who've studied calculus. For those of you who haven't, an integral is just the area under a curve. And you can watch the calculus videos if you want to learn a little bit more about how to do them. And this also applies to the discrete probability distributions. Let me draw one. The sum of all of the probabilities have to be equal to 1. And that example with the dice-- or let's say, since it's faster to draw, the coin-- the two probabilities have to be equal to 1. So this is 1, 0, where x is equal to 1 if we're heads or 0 if we're tails. Each of these have to be 0.5. Or they don't have to be 0.5, but if one was 0.6, the other would have to be 0.4. They have to add to 1. If one of these was-- you can't have a 60% probability of getting a heads and then a 60% probability of getting a tails as well. Because then you would have essentially 120% probability of either of the outcomes happening, which makes no sense at all. So it's important to realize that a probability distribution function, in this case for a discrete random variable, they all have to add up to 1. So 0.5 plus 0.5. And in this case the area under the probability density function also has to be equal to 1. Anyway, I'm all the time for now. In the next video I'll introduce you to the idea of an expected value. See you soon.

Example

Suppose bacteria of a certain species typically live 20 to 30 hours. The probability that a bacterium lives exactly 5 hours is equal to zero. A lot of bacteria live for approximately 5 hours, but there is no chance that any given bacterium dies at exactly 5.00... hours. However, the probability that the bacterium dies between 5 hours and 5.01 hours is quantifiable. Suppose the answer is 0.02 (i.e., 2%). Then, the probability that the bacterium dies between 5 hours and 5.001 hours should be about 0.087, since this time interval is one-tenth as long as the previous. The probability that the bacterium dies between 5 hours and 5.0001 hours should be about 0.0087, and so on.

In this example, the ratio (probability of living during an interval) / (duration of the interval) is approximately constant, and equal to 2 per hour (or 2 hour⁻¹). For example, there is 0.02 probability of dying in the 0.01-hour interval between 5 and 5.01 hours, and (0.02 probability / 0.01 hours) = 2 hour⁻¹. This quantity 2 hour⁻¹ is called the probability density for dying at around 5 hours. Therefore, the probability that the bacterium dies at 5 hours can be written as (2 hour⁻¹) dt. This is the probability that the bacterium dies within an infinitesimal window of time around 5 hours, where dt is the duration of this window. For example, the probability that it lives longer than 5 hours, but shorter than (5 hours + 1 nanosecond), is (2 hour⁻¹)×(1 nanosecond) ≈ 6×10⁻¹³ (using the unit conversion 3.6×10¹² nanoseconds = 1 hour).

There is a probability density function f with f(5 hours) = 2 hour⁻¹. The integral of f over any window of time (not only infinitesimal windows but also large windows) is the probability that the bacterium dies in that window.

Absolutely continuous univariate distributions

A probability density function is most commonly associated with absolutely continuous univariate distributions. A random variable $X$ has density $f_{X}$ , where $f_{X}$ is a non-negative Lebesgue-integrable function, if:

\Pr[a\leq X\leq b]=\int _{a}^{b}f_{X}(x)\,dx.

Hence, if $F_{X}$ is the cumulative distribution function of $X$ , then:

F_{X}(x)=\int _{-\infty }^{x}f_{X}(u)\,du,

and (if

f_{X}

is continuous at

x

)

f_{X}(x)={\frac {d}{dx}}F_{X}(x).

Intuitively, one can think of $f_{X}(x)\,dx$ as being the probability of $X$ falling within the infinitesimal interval $[x,x+dx]$ .

Formal definition

(This definition may be extended to any probability distribution using the measure-theoretic definition of probability.)

A random variable $X$ with values in a measurable space $({\mathcal {X}},{\mathcal {A}})$ (usually $\mathbb {R} ^{n}$ with the Borel sets as measurable subsets) has as probability distribution the measure X_∗P on $({\mathcal {X}},{\mathcal {A}})$ : the density of $X$ with respect to a reference measure $\mu$ on $({\mathcal {X}},{\mathcal {A}})$ is the Radon–Nikodym derivative:

f={\frac {dX_{*}P}{d\mu }}.

That is, f is any measurable function with the property that:

\Pr[X\in A]=\int _{X^{-1}A}\,dP=\int _{A}f\,d\mu

for any measurable set

A\in {\mathcal {A}}.

Discussion

In the continuous univariate case above, the reference measure is the Lebesgue measure. The probability mass function of a discrete random variable is the density with respect to the counting measure over the sample space (usually the set of integers, or some subset thereof).

It is not possible to define a density with reference to an arbitrary measure (e.g. one can not choose the counting measure as a reference for a continuous random variable). Furthermore, when it does exist, the density is almost unique, meaning that any two such densities coincide almost everywhere.

Further details

Unlike a probability, a probability density function can take on values greater than one; for example, the continuous uniform distribution on the interval $[0, 1/2]$ has probability density $f (x) = 2$ for $0 \leq x \leq 1/2$ and $f (x) = 0$ elsewhere.

The standard normal distribution has probability density

f(x)={\frac {1}{\sqrt {2\pi }}}\,e^{-x^{2}/2}.

If a random variable $X$ is given and its distribution admits a probability density function $f$ , then the expected value of $X$ (if the expected value exists) can be calculated as

\operatorname {E} [X]=\int _{-\infty }^{\infty }x\,f(x)\,dx.

Not every probability distribution has a density function: the distributions of discrete random variables do not; nor does the Cantor distribution, even though it has no discrete component, i.e., does not assign positive probability to any individual point.

A distribution has a density function if and only if its cumulative distribution function $F (x)$ is absolutely continuous. In this case: $F$ is almost everywhere differentiable, and its derivative can be used as probability density:

{\frac {d}{dx}}F(x)=f(x).

If a probability distribution admits a density, then the probability of every one-point set ${a}$ is zero; the same holds for finite and countable sets.

Two probability densities $f$ and $g$ represent the same probability distribution precisely if they differ only on a set of Lebesgue measure zero.

In the field of statistical physics, a non-formal reformulation of the relation above between the derivative of the cumulative distribution function and the probability density function is generally used as the definition of the probability density function. This alternate definition is the following:

If $dt$ is an infinitely small number, the probability that $X$ is included within the interval $(t, t + dt)$ is equal to $f (t) dt$ , or:

\Pr(t<X<t+dt)=f(t)\,dt.

Link between discrete and continuous distributions

It is possible to represent certain discrete random variables as well as random variables involving both a continuous and a discrete part with a generalized probability density function using the Dirac delta function. (This is not possible with a probability density function in the sense defined above, it may be done with a distribution.) For example, consider a binary discrete random variable having the Rademacher distribution—that is, taking −1 or 1 for values, with probability 1⁄2 each. The density of probability associated with this variable is:

f(t)={\frac {1}{2}}(\delta (t+1)+\delta (t-1)).

More generally, if a discrete variable can take $n$ different values among real numbers, then the associated probability density function is:

f(t)=\sum _{i=1}^{n}p_{i}\,\delta (t-x_{i}),

where

x_{1},\ldots ,x_{n}

are the discrete values accessible to the variable and

p_{1},\ldots ,p_{n}

are the probabilities associated with these values.

This substantially unifies the treatment of discrete and continuous probability distributions. The above expression allows for determining statistical characteristics of such a discrete variable (such as the mean, variance, and kurtosis), starting from the formulas given for a continuous distribution of the probability.

Families of densities

It is common for probability density functions (and probability mass functions) to be parametrized—that is, to be characterized by unspecified parameters. For example, the normal distribution is parametrized in terms of the mean and the variance, denoted by $\mu$ and $\sigma ^{2}$ respectively, giving the family of densities

f(x;\mu ,\sigma ^{2})={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}.

Different values of the parameters describe different distributions of different random variables on the same sample space (the same set of all possible values of the variable); this sample space is the domain of the family of random variables that this family of distributions describes. A given set of parameters describes a single distribution within the family sharing the functional form of the density. From the perspective of a given distribution, the parameters are constants, and terms in a density function that contain only parameters, but not variables, are part of the normalization factor of a distribution (the multiplicative factor that ensures that the area under the density—the probability of something in the domain occurring— equals 1). This normalization factor is outside the kernel of the distribution.

Since the parameters are constants, reparametrizing a density in terms of different parameters to give a characterization of a different random variable in the family, means simply substituting the new parameter values into the formula in place of the old ones.

Densities associated with multiple variables

For continuous random variables $X 1, ..., X n$ , it is also possible to define a probability density function associated to the set as a whole, often called joint probability density function. This density function is defined as a function of the $n$ variables, such that, for any domain $D$ in the $n$ -dimensional space of the values of the variables $X 1, ..., X n$ , the probability that a realisation of the set variables falls inside the domain $D$ is

\Pr \left(X_{1},\ldots ,X_{n}\in D\right)=\int _{D}f_{X_{1},\ldots ,X_{n}}(x_{1},\ldots ,x_{n})\,dx_{1}\cdots dx_{n}.

If $F (x 1, ..., x n) = Pr(X 1 \leq x 1, ..., X n \leq x n)$ is the cumulative distribution function of the vector $(X 1, ..., X n)$ , then the joint probability density function can be computed as a partial derivative

f(x)=\left.{\frac {\partial ^{n}F}{\partial x_{1}\cdots \partial x_{n}}}\right|_{x}

Marginal densities

For $i = 1, 2, ..., n$ , let $f X i (x i)$ be the probability density function associated with variable $X i$ alone. This is called the marginal density function, and can be deduced from the probability density associated with the random variables $X 1, ..., X n$ by integrating over all values of the other $n - 1$ variables:

f_{X_{i}}(x_{i})=\int f(x_{1},\ldots ,x_{n})\,dx_{1}\cdots dx_{i-1}\,dx_{i+1}\cdots dx_{n}.

Independence

Continuous random variables $X 1, ..., X n$ admitting a joint density are all independent from each other if and only if

f_{X_{1},\ldots ,X_{n}}(x_{1},\ldots ,x_{n})=f_{X_{1}}(x_{1})\cdots f_{X_{n}}(x_{n}).

Corollary

If the joint probability density function of a vector of $n$ random variables can be factored into a product of $n$ functions of one variable

f_{X_{1},\ldots ,X_{n}}(x_{1},\ldots ,x_{n})=f_{1}(x_{1})\cdots f_{n}(x_{n}),

(where each

f i

is not necessarily a density) then the

n

variables in the set are all independent from each other, and the marginal probability density function of each of them is given by

f_{X_{i}}(x_{i})={\frac {f_{i}(x_{i})}{\int f_{i}(x)\,dx}}.

Example

This elementary example illustrates the above definition of multidimensional probability density functions in the simple case of a function of a set of two variables. Let us call ${\vec {R}}$ a 2-dimensional random vector of coordinates $(X, Y)$ : the probability to obtain ${\vec {R}}$ in the quarter plane of positive $x$ and $y$ is

\Pr \left(X>0,Y>0\right)=\int _{0}^{\infty }\int _{0}^{\infty }f_{X,Y}(x,y)\,dx\,dy.

Function of random variables and change of variables in the probability density function

If the probability density function of a random variable (or vector) $X$ is given as $f X (x)$ , it is possible (but often not necessary; see below) to calculate the probability density function of some variable $Y = g (X)$ . This is also called a "change of variable" and is in practice used to generate a random variable of arbitrary shape $f g (X) = f Y$ using a known (for instance, uniform) random number generator.

It is tempting to think that in order to find the expected value $E(g (X))$ , one must first find the probability density $f g (X)$ of the new random variable $Y = g (X)$ . However, rather than computing

\operatorname {E} {\big (}g(X){\big )}=\int _{-\infty }^{\infty }yf_{g(X)}(y)\,dy,

one may find instead

\operatorname {E} {\big (}g(X){\big )}=\int _{-\infty }^{\infty }g(x)f_{X}(x)\,dx.

The values of the two integrals are the same in all cases in which both $X$ and $g (X)$ actually have probability density functions. It is not necessary that $g$ be a one-to-one function. In some cases the latter integral is computed much more easily than the former. See Law of the unconscious statistician.

Scalar to scalar

Let $g:\mathbb {R} \to \mathbb {R}$ be a monotonic function, then the resulting density function is^[5]

f_{Y}(y)=f_{X}{\big (}g^{-1}(y){\big )}\left|{\frac {d}{dy}}{\big (}g^{-1}(y){\big )}\right|.

Here $g -1$ denotes the inverse function.

This follows from the fact that the probability contained in a differential area must be invariant under change of variables. That is,

\left|f_{Y}(y)\,dy\right|=\left|f_{X}(x)\,dx\right|,

f_{Y}(y)=\left|{\frac {dx}{dy}}\right|f_{X}(x)=\left|{\frac {d}{dy}}(x)\right|f_{X}(x)=\left|{\frac {d}{dy}}{\big (}g^{-1}(y){\big )}\right|f_{X}{\big (}g^{-1}(y){\big )}={\left|\left(g^{-1}\right)'(y)\right|}\cdot f_{X}{\big (}g^{-1}(y){\big )}.

For functions that are not monotonic, the probability density function for $y$ is

\sum _{k=1}^{n(y)}\left|{\frac {d}{dy}}g_{k}^{-1}(y)\right|\cdot f_{X}{\big (}g_{k}^{-1}(y){\big )},

where

n (y)

is the number of solutions in

x

for the equation

g(x)=y

, and

g_{k}^{-1}(y)

are these solutions.

Vector to vector

Suppose $x$ is an $n$ -dimensional random variable with joint density $f$ . If $y = G (x)$ , where $G$ is a bijective, differentiable function, then $y$ has density $p Y$ :

p_{Y}(\mathbf {y} )=f{\Bigl (}G^{-1}(\mathbf {y} ){\Bigr )}\left|\det \left[\left.{\frac {dG^{-1}(\mathbf {z} )}{d\mathbf {z} }}\right|_{\mathbf {z} =\mathbf {y} }\right]\right|

with the differential regarded as the Jacobian of the inverse of

G (\cdot)

, evaluated at

y

.^[6]

For example, in the 2-dimensional case $x = (x 1, x 2)$ , suppose the transform $G$ is given as $y 1 = G 1 (x 1, x 2)$ , $y 2 = G 2 (x 1, x 2)$ with inverses $x 1 = G 1 -1 (y 1, y 2)$ , $x 2 = G 2 -1 (y 1, y 2)$ . The joint distribution for y = (y₁, y₂) has density^[7]

p_{Y_{1},Y_{2}}(y_{1},y_{2})=f_{X_{1},X_{2}}{\big (}G_{1}^{-1}(y_{1},y_{2}),G_{2}^{-1}(y_{1},y_{2}){\big )}\left\vert {\frac {\partial G_{1}^{-1}}{\partial y_{1}}}{\frac {\partial G_{2}^{-1}}{\partial y_{2}}}-{\frac {\partial G_{1}^{-1}}{\partial y_{2}}}{\frac {\partial G_{2}^{-1}}{\partial y_{1}}}\right\vert .

Vector to scalar

Let $V:\mathbb {R} ^{n}\to \mathbb {R}$ be a differentiable function and $X$ be a random vector taking values in $\mathbb {R} ^{n}$ , $f_{X}$ be the probability density function of $X$ and $\delta (\cdot )$ be the Dirac delta function. It is possible to use the formulas above to determine $f_{Y}$ , the probability density function of $Y=V(X)$ , which will be given by

f_{Y}(y)=\int _{\mathbb {R} ^{n}}f_{X}(\mathbf {x} )\delta {\big (}y-V(\mathbf {x} ){\big )}\,d\mathbf {x} .

This result leads to the law of the unconscious statistician:

\operatorname {E} _{Y}[Y]=\int _{\mathbb {R} }yf_{Y}(y)\,dy=\int _{\mathbb {R} }y\int _{\mathbb {R} ^{n}}f_{X}(\mathbf {x} )\delta {\big (}y-V(\mathbf {x} ){\big )}\,d\mathbf {x} \,dy=\int _{{\mathbb {R} }^{n}}\int _{\mathbb {R} }yf_{X}(\mathbf {x} )\delta {\big (}y-V(\mathbf {x} ){\big )}\,dy\,d\mathbf {x} =\int _{\mathbb {R} ^{n}}V(\mathbf {x} )f_{X}(\mathbf {x} )\,d\mathbf {x} =\operatorname {E} _{X}[V(X)].

Proof:

Let $Z$ be a collapsed random variable with probability density function $p_{Z}(z)=\delta (z)$ (i.e., a constant equal to zero). Let the random vector ${\tilde {X}}$ and the transform $H$ be defined as

H(Z,X)={\begin{bmatrix}Z+V(X)\\X\end{bmatrix}}={\begin{bmatrix}Y\\{\tilde {X}}\end{bmatrix}}.

It is clear that $H$ is a bijective mapping, and the Jacobian of $H^{-1}$ is given by:

{\frac {dH^{-1}(y,{\tilde {\mathbf {x} }})}{dy\,d{\tilde {\mathbf {x} }}}}={\begin{bmatrix}1&-{\frac {dV({\tilde {\mathbf {x} }})}{d{\tilde {\mathbf {x} }}}}\\\mathbf {0} _{n\times 1}&\mathbf {I} _{n\times n}\end{bmatrix}},

which is an upper triangular matrix with ones on the main diagonal, therefore its determinant is 1. Applying the change of variable theorem from the previous section we obtain that

f_{Y,X}(y,x)=f_{X}(\mathbf {x} )\delta {\big (}y-V(\mathbf {x} ){\big )},

which if marginalized over

x

leads to the desired probability density function.

Sums of independent random variables

The probability density function of the sum of two independent random variables $U$ and $V$ , each of which has a probability density function, is the convolution of their separate density functions:

f_{U+V}(x)=\int _{-\infty }^{x}f_{U}(y)f_{V}(x-y)\,dy=\left(f_{U}*f_{V}\right)(x)

It is possible to generalize the previous relation to a sum of N independent random variables, with densities $U 1, ..., U N$ :

f_{U_{1}+\cdots +U_{N}}(x)=\left(f_{U_{1}}*\cdots *f_{U_{N}}\right)(x)

This can be derived from a two-way change of variables involving $Y = U + V$ and $Z = V$ , similarly to the example below for the quotient of independent random variables.

Products and quotients of independent random variables

Given two independent random variables $U$ and $V$ , each of which has a probability density function, the density of the product $Y = UV$ and quotient $Y = U / V$ can be computed by a change of variables.

Example: Quotient distribution

To compute the quotient $Y = U / V$ of two independent random variables $U$ and $V$ , define the following transformation:

{\begin{aligned}Y&=U/V\\[1ex]Z&=V\end{aligned}}

Then, the joint density $p (y, z)$ can be computed by a change of variables from U,V to Y,Z, and $Y$ can be derived by marginalizing out $Z$ from the joint density.

The inverse transformation is

{\begin{aligned}U&=YZ\\V&=Z\end{aligned}}

The absolute value of the Jacobian matrix determinant $J(U,V\mid Y,Z)$ of this transformation is:

\left|\det {\begin{bmatrix}{\frac {\partial u}{\partial y}}&{\frac {\partial u}{\partial z}}\\{\frac {\partial v}{\partial y}}&{\frac {\partial v}{\partial z}}\end{bmatrix}}\right|=\left|\det {\begin{bmatrix}z&y\\0&1\end{bmatrix}}\right|=|z|.

Thus:

p(y,z)=p(u,v)\,J(u,v\mid y,z)=p(u)\,p(v)\,J(u,v\mid y,z)=p_{U}(yz)\,p_{V}(z)\,|z|.

And the distribution of $Y$ can be computed by marginalizing out $Z$ :

p(y)=\int _{-\infty }^{\infty }p_{U}(yz)\,p_{V}(z)\,|z|\,dz

This method crucially requires that the transformation from U,V to Y,Z be bijective. The above transformation meets this because $Z$ can be mapped directly back to $V$ , and for a given $V$ the quotient $U / V$ is monotonic. This is similarly the case for the sum $U + V$ , difference $U - V$ and product $UV$ .

Exactly the same method can be used to compute the distribution of other functions of multiple independent random variables.

Example: Quotient of two standard normals

Given two standard normal variables $U$ and $V$ , the quotient can be computed as follows. First, the variables have the following density functions:

{\begin{aligned}p(u)&={\frac {1}{\sqrt {2\pi }}}e^{-{u^{2}}/{2}}\\[1ex]p(v)&={\frac {1}{\sqrt {2\pi }}}e^{-{v^{2}}/{2}}\end{aligned}}

We transform as described above:

{\begin{aligned}Y&=U/V\\[1ex]Z&=V\end{aligned}}

This leads to:

{\begin{aligned}p(y)&=\int _{-\infty }^{\infty }p_{U}(yz)\,p_{V}(z)\,|z|\,dz\\[5pt]&=\int _{-\infty }^{\infty }{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}y^{2}z^{2}}{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {1}{2}}z^{2}}|z|\,dz\\[5pt]&=\int _{-\infty }^{\infty }{\frac {1}{2\pi }}e^{-{\frac {1}{2}}\left(y^{2}+1\right)z^{2}}|z|\,dz\\[5pt]&=2\int _{0}^{\infty }{\frac {1}{2\pi }}e^{-{\frac {1}{2}}\left(y^{2}+1\right)z^{2}}z\,dz\\[5pt]&=\int _{0}^{\infty }{\frac {1}{\pi }}e^{-\left(y^{2}+1\right)u}\,du&&u={\tfrac {1}{2}}z^{2}\\[5pt]&=\left.-{\frac {1}{\pi \left(y^{2}+1\right)}}e^{-\left(y^{2}+1\right)u}\right|_{u=0}^{\infty }\\[5pt]&={\frac {1}{\pi \left(y^{2}+1\right)}}\end{aligned}}

This is the density of a standard Cauchy distribution.

References

^ "AP Statistics Review - Density Curves and the Normal Distributions". Archived from the original on 2 April 2015. Retrieved 16 March 2015.
^ Grinstead, Charles M.; Snell, J. Laurie (2009). "Conditional Probability - Discrete Conditional" (PDF). Grinstead & Snell's Introduction to Probability. Orange Grove Texts. ISBN 978-1616100469. Archived (PDF) from the original on 2003-04-25. Retrieved 2019-07-25.
^ "probability - Is a uniformly random number over the real line a valid distribution?". Cross Validated. Retrieved 2021-10-06.
^ Ord, J.K. (1972) Families of Frequency Distributions, Griffin. ISBN 0-85264-137-0 (for example, Table 5.1 and Example 5.4)
^ Siegrist, Kyle. "Transformations of Random Variables". LibreTexts Statistics. Retrieved 22 December 2023.
^ Devore, Jay L.; Berk, Kenneth N. (2007). Modern Mathematical Statistics with Applications. Cengage. p. 263. ISBN 978-0-534-40473-4.
^ David, Stirzaker (2007-01-01). Elementary Probability. Cambridge University Press. ISBN 978-0521534284. OCLC 851313783.

External links

Ushakov, N.G. (2001) [1994], "Density of a probability distribution", Encyclopedia of Mathematics, EMS Press
Weisstein, Eric W. "Probability density function". MathWorld.

v t e Theory of probability distributions
probability mass function (pmf) probability density function (pdf) cumulative distribution function (cdf) quantile function
raw moment central moment mean variance standard deviation skewness kurtosis L-moment
moment-generating function (mgf) characteristic function probability-generating function (pgf) cumulant combinant

This page was last edited on 9 April 2024, at 01:15

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Transcription

Example

Absolutely continuous univariate distributions

Formal definition

Discussion

Further details

Link between discrete and continuous distributions

Families of densities

Densities associated with multiple variables

Marginal densities

Independence

Corollary

Example

Function of random variables and change of variables in the probability density function

Scalar to scalar

Vector to vector

Vector to scalar

Sums of independent random variables

Products and quotients of independent random variables

Example: Quotient distribution

Example: Quotient of two standard normals

See also

References

Further reading

External links