Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a random variable J is joint distributed between X and Y. Then,

$$ E(J) = \int_{-\infty}^\infty \int_{-\infty}^\infty f(x, y) dxdy $$

However, how do I calculate the variance of $J$? Had $J$ been a non-jointly distributed random variable, then we can use $Var(J) = E(J^2) - E(J)^2$. However, what is $E(J^2)$ in this case?

By the law of the unconscious statistician, $E(J^2) = \int_{-\infty}^\infty \int_{-\infty}^\infty j^2 f(x, y) dxdy$.

What should I supply for $j^2$ though? $x^2$ or $y^2$?

share|improve this question
    
Presumably $f(x,y)$ is the joint density function. Then our double integral is $1$. For the expectation of $J$, we need to know something about $J$. Maybe $J=g(X,Y)$? Then we can find the mean and variance of $J$. –  André Nicolas Nov 4 '12 at 4:52
    
Thanks! $f(x, y)$ is indeed the joint density function. What is $g(x, y)$ though? How does that differ from the joint density function? What if $X$ and $Y$ are independent? –  David Faux Nov 4 '12 at 4:54
    
You presumably want to find the variance of some random variable. I assumed that you were calling that random variable $J$. If we know $J$ as a function of $X$ and $Y$, we can do the calculation. So to be specific, what do you want to find the variance of? It has to be a random variable. –  André Nicolas Nov 4 '12 at 4:57
    
Specifically, the variance of $J$. –  David Faux Nov 4 '12 at 4:58
    
Like say $J$ is the minimum of $X$ and $Y$ ... –  David Faux Nov 4 '12 at 5:04
show 1 more comment

2 Answers 2

up vote 3 down vote accepted

The question has evolved in the comments, and now may be asking the following. Let $J$ be the minimum of $X$ and $Y$. What is the variance of $J$? We assume that the joint density function of $X$ and $Y$ is $f(x,y)$.

An answer goes as follows. Let $m(x,y)$ be the minimum of $x$ and $y$. Then $$E(J)=\int_{-\infty}^\infty\int_{-\infty}^\infty m(x,y)f(x,y)\,dx\,dy.$$ As for the variance, it is $E(J^2)-(E(J))^2$, and $$E(J^2)=\int_{-\infty}^\infty\int_{-\infty}^\infty (m(x,y))^2f(x,y)\,dx\,dy.$$

In evaluating the integrals, we probably will want to use the following strategy, which we illustrate with the integral for the mean of $J$. Divide the plane into two parts, the part below $y=x$ and the part above. Then our integral is the sum of the integrals over the two parts.

In the part with $y\lt x$, we have $m(x,y)=y$. So our integral over this part is $$\int_{x=-\infty}^\infty\int_{y=-\infty}^x yf(x,y)\,dy\,dx.$$ The integral over the part where $x\lt y$ is obtained in the same way, except for some minor changes. It is $$\int_{y=-\infty}^\infty\int_{x=-\infty}^y xf(x,y)\,dx\,dy.$$ Add these.

The integral for calculating $E(J^2)$ can be broken up in exactly the same way. Instead of integrating $yf(x,y)$ or $xf(x,y)$ over suitable regions, we will be integrating $y^2f(x,y)$ and $x^2f(x,y)$ over the same regions.

share|improve this answer
    
Thanks for being so thorough! –  David Faux Nov 4 '12 at 5:59
add comment

Let $\bar F:(x,y)\mapsto\mathbb P(X\geqslant x,Y\geqslant y)$ denote the complementary CDF of $(X,Y)$ and $G=-(\partial_x\bar F+\partial_y\bar F)$. Then $t\mapsto G(t,t)$ is the PDF of $J=\min\{X,Y\}$ hence $$ \mathbb E(J)=\int tG(t,t)\mathrm dt,\qquad\mathbb E(J^2)=\int t^2G(t,t)\mathrm dt, $$ from which the variance of $J$ follows.

In the case when $(X,Y)$ is independent, $\bar F(x,y)=\bar F_X(x)\bar F_Y(y)$ where $\bar F_X$ and $\bar F_Y$ are the complementary CDF of $X$ and $Y$ respectively, defined by $\bar F_X(x)=\mathbb P(X\geqslant x)$ and $\bar F_Y(y)=\mathbb P(Y\geqslant y)$. If furthermore the distributions of $X$ and $Y$ have densities $f_X$ and $f_Y$, then $\partial_x\bar F_X(x)=-f_X(x)$ and $\partial_y\bar F_Y(y)=-f_Y(y)$ hence $$ G(x,y)=f_X(x)\bar F_Y(y)+f_Y(y)\bar F_X(x), $$ and the formulas above apply, with $$ G(t,t)=f_X(t)\bar F_Y(t)+f_Y(t)\bar F_X(t)=f_X(t)\mathbb P(Y\geqslant t)+f_Y(t)\mathbb P(X\geqslant t). $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.