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When I need to find $E(X|Y=1)$ and have: $$f(x,y)=\begin{cases}\lambda^3xe^{-\lambda y} & 0\lt x\lt y\\ 0 & \text{elsewhere}\end{cases}$$ Do I need to find both marginal densities to do this? I know that the definition of $E(X|Y=y)=$$$\begin{align}\int_{-\infty}^{\infty} g(x)f_X(x|Y=1)dx\\ =\int_{-\infty}^{\infty}g(x)P(x|Y=1)dx\\=\int_{-\infty}^{\infty}g(x)*\frac{f(x,y)}{f_Y(y)} dx\end{align}$$ This is the definitiion given in my textbook. Is the $g(x)$ supposed to be the marginal density of $X$?

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2 Answers 2

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First you need to find $f_Y(y) = \int_{-\infty}^{\infty}f(x,y)dx$.

Then, $$E(X|Y=y) = \int_{-\infty}^{\infty}x\frac{f(x,y)}{f_Y(y)}dx$$ The formula you have written is correct for $E(g(X)|Y=y)$, which is a generalized version of the above.

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So when we plug in for $y=1$ would the upper limit on the integral be $1$ since it's saying $y=1$? –  TheHopefulActuary Dec 13 '12 at 4:04
yeah. you may do so in this case. –  dexter04 Dec 13 '12 at 4:07
What kind of cases wouldn't that be acceptable? –  TheHopefulActuary Dec 13 '12 at 4:10
It is OK to change the limits to the range where $f(x,y)$ is non-zero. –  dexter04 Dec 13 '12 at 7:21

I think your book is telling you $E[g(X)\mid Y = y]$ for arbitrary measurable function $g(\cdot)$, not $E[X\mid Y = y]$. And yes, you do need to compute the marginal density of $Y$ in order to divide $f_{X,Y}(x,y)$ in your formula. But you don't need to find both marginal densities, and in this simple case, it is even unnecessary to compute $f_Y(y)$ explicitly; one can get find $f_{X\mid Y=1}$ without knowing the formula for $f_Y(y)$ or what kind of random variable $Y$ is (it is a Gamma random variable, by the way).

The way I would approach the problem is to say that given $Y = 1$, the conditional density of $X$ has shape given by $f(x,1)$ which is a linear function of $x$ for $x \in (0,1)$. The graph shows a triangle whose area must equal $1$ in order to get a valid density. So the conditional density must be $$f_{X\mid Y=1}(x\mid Y=1) = 2x\mathbf 1_{(0,1)}$$ and so the mean is $\frac{2}{3}$.

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And you got the $f_{X|Y=1}(x|Y=1)=2x1_{(0,1)}$ from the density function divided by $Y$'s marginal density correct? –  TheHopefulActuary Dec 13 '12 at 3:52
Well, I didn't actually bother to compute $f_Y(y)$ explicitly. Instead, I claim that the conditional density of $X$ given $Y = y_0$ is proportional to $$f_{X,Y}(x,y_0) = (\lambda^3\exp(-\lambda y_0))\cdot x\mathbf 1_{(0,y_0)} = c\cdot x\mathbf 1_{(0,y_0)}$$ whose graph is a straight line with value $0$ at $x=0$ and value $c\cdot y_0$ as $x\to y_0$. The area under this curve is $\frac{1}{2}(c\cdot y_0)y_0$, and so it must be that $$f_{X\mid Y=y_0}(x\mid Y=y_0) = \frac{2}{y_0^2}x\mathbf 1_{(0,y_0)}.$$ –  Dilip Sarwate Dec 13 '12 at 11:58

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