let $X$ and $Y$ be two exponential random variables with the same parameter $\lambda$. I try to calculate the MGF of random variable $V=X-Y$ given that $X>Y$. We know that the MGF of Exponential random variables are $M_X(t)=M_Y(t)=\frac{\lambda}{\lambda-t}$. Further, because of the memoryless property of exponential random variables, if $X$ and $Y$ are independent, then $V$ is an exponential random variable with parameter $\lambda$. However, I try to calculate the MGF of $V$ directly as follows:
$$E[e^{tV}|X>Y]=\int_0^\infty E[e^{t(X-Y)}|X>Y=y]dF_Y(y)$$
$$E[e^{t(X-Y)}|X>Y=y]=\int_y^\infty e^{t(X-y)}\lambda e^{-\lambda x} dx=\frac{\lambda e^{-\lambda y}}{\lambda-t}$$
therefore:
$$M_V(t)=E[e^{tV}|X>Y]=\int_0^\infty \frac{\lambda e^{-\lambda y}}{\lambda-t}dF_Y(y)=\int_0^\infty \frac{\lambda e^{-\lambda y}}{\lambda-t}\lambda e^{-\lambda y} dy=\frac{1}{2}\frac{\lambda}{\lambda-t}$$
i dont know why the final result is different from $M_X(t)$ by factor of$\frac{1}{2}$
Best Answer
This is incorrect: $$E[e^{t(X-Y)}|X>Y=y]=\int_y^\infty e^{t(X-y)}\lambda e^{-\lambda x} dx$$
The reason is becuase $$\operatorname{E}[g(X) \mid X > y] \ne \int_{x=y}^\infty g(x) f_X(x) \, dx,$$ because $$\int_{x=y}^\infty f_X(x) \, dx \ne 1.$$ The expectation must always be calculated over a proper density that integrates to $1$.
Instead, $$\operatorname{E}[e^{t(X-y)} \mid X > y] = \int_{x=y}^\infty e^{t(x-y)} \lambda e^{-\lambda(x\color{red}{-y})} \, dx.$$ This however, still implicitly uses the memoryless property as justification. You could also write
$$\operatorname{E}[g(X) \mid X > y] = \frac{\int_{x=y}^\infty g(x) f_X(x) \, dx}{\int_{x=y}^\infty f_X(x) \, dx}.$$ Here, the conditional distribution is $X \mid X > y$ and has density $$f_{X \mid X > y}(x) = \frac{f_X(x)}{S_X(y)}$$ where $S_X = 1 - F_X$ is the survival function.