# conditional expectation — simplify

Let $x$ and $y$ be two independent random random variables with densities $f(x)$ and $f(y)$. I intend to define $\int \int_{-\infty}^x f(x)f(y)\,dy\,dx$ in relation to $E[x\mid x>y]$. I attempted the following. $$E[x\mid x>y]=\frac{\int \int_{-\infty}^{x} xf(x)f(y)\,dy\,dx}{\int \int_{-\infty}^x f(x)f(y)\,dy\,dx}=\frac{\int xF_{y}(x)f(x)\,dx}{\int F_{y}(x)f(x)\,dx}$$ where $F$ is CDF. I am not sure if this is correct, though. Can somebody please tell me if I am right or suggest me how to correct if I made a mistake.

• Can we call $f(y)$ another name, like $g(y)$ or at least ($f_x(x)$ and $f_y(y)$)? – Berci Nov 30 '12 at 11:08
• @Berci I made that on purpose as $x$ and $y$ are identical and independently distributed (I forgot to mention it). Yet, if my notation appears to be ambiguous we can use yours. – Daniel Lårs Nov 30 '12 at 11:24
• Once you will have renamed $X$ and $Y$ (instead of $x$ and $y$) the i.i.d. random variables and $F$ (instead of $F_y$) the CDF, the formula shall be correct. – Did Nov 30 '12 at 11:39