conditional-expectationprobabilityrandom variables
I was wondering if anyone could help me with the following question specifically:
The continuous random variables $X_1$ and $X_2$ have the following joint probability density function: $$f(x_1,x_2)=\frac2{27}$$ over $0< x_1< 3$ and $0< x_2< 9-3x_1.$
Find $$E(X_1\mid X_2=5).$$
I tried to use the method of integrating $xh(x_1\mid x_2)$ but I couldn't get it to work. Thank you in advance!
(1) The marginal PDF of $X$ is $$f_X(x)=\int_{-\infty}^\infty f(x,y)\,dy\\ =\begin{cases}\int_0^x 8xy\,dy & ,\text{if} \,\,\, 0<x<1\\0 & \text{otherwise}\end{cases}$$ The marginal PDF of $Y$ is $$f_Y(y)=\int_{-\infty}^\infty f(x,y)\,dx\\ =\begin{cases}\int_y^1 8xy\,dx & ,\text{if} \,\,\, 0<y<1\\0 & \text{otherwise}\end{cases}$$
(2) The conditional PDF of $Y|X=x$ is $$f_{Y|X}(y|x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}\\ =\begin{cases}\dfrac{8xy}{\int_0^x 8xy\,dy} & ,\text{if} \,\,\, 0<y<x\\0 & \text{otherwise}\end{cases}\\=\begin{cases}\dfrac{2y}{x^2} & ,\text{if} \,\,\, 0<y<x\\0 & \text{otherwise}\end{cases}$$ So $P(Y \le \frac{1}{4} \mid X = \frac{1}{2})=\int_0^{1/4}\dfrac{2y}{(1/2)^2}\,dy$
If I were you, I would use transformations.
$$ Y1 = X1 - X2 = u_1(x_1,x_2)$$ $$ Y2 = X1 + X2 = u_2(x_1,x_2)$$ $$ X_1 = \frac{Y_1+Y_2}{2}=w_1(y_1,y_2)$$ $$ X_2 = \frac{Y_1-Y_2}{2}=w_2(y_1,y_2)$$
$$ f(y_1,y_2) = f(w_1(y1, y2),w_2(y1,y2)|J|$$
j is jaconian
I solved same problem with standard normal distributions. I added picture of my process. or You can solve this problem by using distribution method.
good luck :)
Best Answer
Detailed calculation using no intuition
The conditional expectation of $X_1$ given that $X_2=5$ can be calculated if we know the corresponding conditional density, $f_{X_2\mid X_1=5}(x_2).$ This conditional density can be calculated as follows:
$$f_{X_1\mid X_2=5}(x_1)=\frac{f_{X_1,X_2}(x_1,5)}{f_{X_2}(5)}=\frac{\frac2{27}}{f_{X_2}(5)}$$ where $f_{X_2}(5)$ is the marginal pdf of $X_2$ at $x_2=5$.
Now, the marginal density of $X_2$ is the integral of the common density with respect to $x_1$:
$$f_{X_2}(x_2)=\int_0^{3-\frac13x_2}f(x_1,x_2)\ dx_1=\frac2{27}\left(3-\frac13x_2\right)$$ if $0\leq x_2\leq 9$ So, $$f_{X_2}(5)=\frac8{3^4}.$$
That is $$f_{X_1\mid X_2=5}(x_1)=\frac34$$ if $0\leq x_1\leq \frac43$.
Indeed, this is uniform over $[0,\frac43]$ . The conditional expectation is then $$\frac23$$ (half of $\frac43$) "because"
$$\frac34\int_0^{\frac43}x_1\ dx_1=\frac23.$$