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.

Given two random variables, X and Y with joint density $$ f(x,y)= c*cosx$$for $0<y<x<\pi/2$ (and zero otherwise),

how do you find the conditional expectation $E(Y|X=x)$?

A general method is welcome. I'm just not sure where to begin.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

Draw the (nonzero density) range of $X,Y$, now it is a triangle (without its boundary) on the plane with vertices $(0,0)$, $(\pi/2,0)$ and $(\pi/2,\pi/2)$. (In integrals $x$ can go from $0$ to $\pi/2$ and $y$ from $0$ to $x$, or if you exchange them, $y=0..\pi/2$ and $x=y..\pi/2$.)

Now for any set $U$ within the triangle, by the definition of (joint) density function, we have that $$P((X,Y)\in U)=\int_U f(x,y)dxdy$$ and so, if $U$ is the whole triangle, we have to get $1$. This will determine $c$.

Now the conditional probability, specially in the case $X=x$ simply restricts the possible ranges, to the vertical segment of the triangle at $X=x$. So, now this $x$ is considered as fixed.

Let $C:=\int_0^x f(x,y)dy$, now it is $x\cdot c\cdot \cos x$, we are going to divide with it and use $y\mapsto \displaystyle\frac{f(x,y)}C$ as the density function of the conditional random variable $Y|X=x$. Its expectational value is $$E(Y|X=x)=\frac1C\int_0^x y\cdot f(x,y)dy.$$

share|improve this answer
    
This was very helpful! following through, I ended up with E(Y|X=x)=x/2, which is the correct answer from the book. –  user62596 Feb 17 '13 at 1:11

Since the event $X=x$ has probability zero, conditioning upon it happening is a tricky business: see Borel-Kolmogorov paradox.

But we can close our eyes, cross our fingers, integrate $yf(x,y)$ over the line with fixed value of $x$, and divide by the integral of $f$ over the same line. This works when $0<x<\pi/2$ and gives a reasonably-looking number.

(And the computation makes rigorous sense, if we think of conditional expectation as a map on a Lebesgue space, so that we focus not on pointwise values but on the density).

share|improve this answer

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.