A question on conditional expectation of a random variable

Question

Consider the joint probability density function:

$$f(x_1,x_2)= \begin{cases} 2e^{-2x_1}, & \text{ for } 0 \le x_2 \le x_1 < \infty \\ 0, & \text{ elsewhere} \\ \end{cases} $$

Find the value of $E(X_1|X_2=x_2 )$ for $0 \le x_2 < \infty$

My attempt:

We know the formula for $$E(X|Y=y)= \int_{-\infty}^{\infty} x f_{X|Y}(x|y)dx=\frac{1}{f_Y(y)}x \int_{-\infty}^{\infty} f_{X,Y}(x,y)dx,$$ where $f_Y(y)$ is the marginal density function of the random variable $Y$.

Applying it here in this case, we have that $f_{X_2}(x_2)=\int_{x_2}^{\infty} 2e^{-2x_1}dx_1=e^{-2x_2}.$

So, we have that $$E(X|Y=y)=\frac{1}{e^{-2x_2}}\int_{x_2}^{\infty} x_12e^{-2x_1}dx_1= 2e^{2x_2} \bigg[x_1 \bigg(\frac{e^{-2x_1}}{-2} \bigg)-\bigg(\frac{e^{-2x_1}}{4}\bigg) \bigg]_{x_2}^{\infty}= 2e^{2x_2}\bigg[x_2 \bigg(\frac{e^{-2x_2}}{2} \bigg)+\bigg(\frac{e^{-2x_2}}{4}\bigg) \bigg] = x_2 + \frac{1}{2}.$$

So, $E(X_1|X_2=x_2 ) = x_2 + \frac{1}{2}.$

I have a problem here.

My answer is not matching with any of the options given.

I want to know where I have gone wrong. Please point out my mistakes, if any.

Thanks in advance.

Answer 1

Your computations are correct as shown, but there is a problem with the joint density, since $$\int_{x_2=0}^\infty \int_{x_1=x_2}^\infty 2e^{-2x_1} \, dx_1 \, dx_2 = \int_{x_2=0}^\infty e^{-2x_2} \, dx_2 = \frac{1}{2} \ne 1.$$ Check that the joint density is correctly specified.