Definition

The joint distribution of two random variables \(X\) and \(Y\) is the function \(F(A) = P((X, Y) \in A)\) where \(A \subset \mathbb{R}^2.\)

Two discrete random variables \(X\) and \(Y\) have a joint probability mass function, or joint pmf, \(p(x, y).\) The joint pmf satisfies the following: \[P((X, Y) \in A) = \sum_{(x,y) \in A} p(x,y)\]

Two continuous random variables \(X\) and \(Y\) have a joint probability density function, or joint pdf, \(f(x, y).\) The joint pdf satisfies the following: \[P((X, Y) \in A) = \int \int_A f(x,y) dx dy\]

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Example: Let \(X\) and \(Y\) be random variables with joint pmf \(p(0,0) = 0.3,\) \(p(1,0) = 0.1,\) \(p(4,5) = 0.1,\) \(p(4,4) = 0.5.\) Find \(P(X = Y).\)

Solution: The two points at which \(X = Y\) are \((0,0)\) and \((4,4).\) So, \begin{align} P(X = Y) & = p(0,0) + p(4,4) \\ & = 0.3 + 0.5 \\ & = 0.8 \end{align}

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Example: Let \(X\) and \(Y\) be random variables with joint pdf \(f(x,y) = 4xy\) if \(0 \leq x \leq 1\) and \(0 \leq y \leq 1,\) and \(f(x,y) = 0\) otherwise. Find \(P(X < Y).\)

Solution: The region on which \(X < Y\) is the pairs of points \((x,y)\) where \(x < y.\) So, \(A = \{(x,y) : x < y\}.\) Integrating over the region where \(f(x,y) \neq 0,\) the integral should be set up as \[\int_0^1 \int_0^y f(x,y) dxdy\] Plugging in \(f(x,y) = 4xy,\) we get \begin{align} \int_0^1 \int_0^y f(x,y) dxdy & = \int_0^1 \int_0^y 4xy dxdy \\ & = \int_0^1 4y\left(\left.\frac{x^2}{2}\right|_0^y\right)y dy \\ & = \int_0^1 2y^3 dy \\ & = 2 \left(\left.\frac{y^4}{4}\right|_0^1\right) \\ & = \frac{1}{2} \end{align}

Marginal Distributions

Given the joint pmf \(p(x,y)\) of two random variables \(X\) and \(Y,\) one can find the pmf's of both \(X\) and \(Y\) by summing the other variable.

The pmf of \(X\) is \[p_X(x) = \sum_y p(x,y)\] The pmf of \(Y\) is \[p_Y(y) = \sum_x p(x,y)\]

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Example: Let \(X\) and \(Y\) be random variables with joint pmf \(p(0,0) = 0.3,\) \(p(1,0) = 0.1,\) \(p(4,5) = 0.1,\) \(p(4,4) = 0.5.\) What is the pmf of \(X\)?

There are \(3\) values that \(X\) can take, \(0,\) \(1\) and \(4.\) For each value of \(X,\) sum over all values \(Y\) can take. \begin{align} & p(0) = p(0,0) = 0.3 \\ & p(1) = p(1,0) = 0.1 \\ & p(4) = p(4,5) + p(4,4) = 0.1 + 0.5 = 0.6 \end{align}

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Given the joint pdf \(f(x,y)\) of two random variables \(X\) and \(Y,\) one can find the pdf's of both \(X\) and \(Y\) by integrating the other variable.

The pdf of \(X\) is \[f_X(x) = \int_{-\infty}^\infty f(x,y) dy\] The pdf of \(Y\) is \[f_Y(y) = \int_{-\infty}^\infty f(x,y) dx\]

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Example: Let \(X\) and \(Y\) have joint pdf \(f(x,y) = \frac{1}{\pi}\) if \(x^2 + y^2 \leq 1.\) This means that \((X,Y)\) are equally likely to be any point on the unit circle. What is the pdf of \(X?\)

Let \(f(x)\) be the pdf of \(X.\) Since the probability is only positive if \(x^2+y^2 \leq 1,\) \(f(x)\) will only be positive if \(-1 \leq x \leq 1.\)

Let \(x\) be a value between \(-1\) and \(1.\) Then \(f(x,y)\) is positive if \(-\sqrt{1-x^2} < y < \sqrt{1-x^2}.\) Using these bounds, we can compute \(f(x).\) \begin{align} f(x) & = \int_{-\infty}^\infty f(x,y)dy \\ & = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy \\ & = \frac{2}{\pi}\sqrt{1-x^2} \end{align} Therefore, the pdf of \(X\) is \(f(x) = \frac{2}{\pi}\sqrt{1-x^2}\) for \(-1 < x < 1\) and \(f(x) = 0\) otherwise.

Multiple Continuous Variables

Given variables \(X_1, X_2, \dots, X_n,\) the joint distribution of all \(n\) variables is \(P((X_1, X_2, \dots, X_n) \in A)\) for some \(A \subset \mathbb{R}^n.\)

If \(X_1, X_2, \dots, X_n\) are continuous random variables, then they will have a joint pdf \(f(x_1, x_2, \dots, x_n)\) such that \[P((X_1, X_2, \dots, X_n) \in A) = \int \int \dots \int_A f(x_1, x_2, \dots, x_n) dx_1 dx_2 \dots dx_n\]