# An inequality about conditional expectation

Let X be an non-negative random variable. Can any one tell me how to prove the following inequality?

$\mathbb{P}(X>\lambda_0)\cdot \mathbb{E}[X|X>\lambda_0]\leq \int_{\lambda_0}^{\infty}\lambda\cdot \mathbb{P}(X>\lambda)d\lambda$.

$\lambda_0$ is a positive number.

I am very confused about the conditional expectation here. Is my following calculation true?

$\mathbb{E}[X|X>\lambda_0]=\int_0^{\infty}\frac{\mathbb{P}(X>\lambda|X>\lambda_0)}{\mathbb{P}(X>\lambda_0)}d\lambda=\int_0^{\lambda_0}d\lambda+\int_{\lambda_0}^{\infty}\frac{\mathbb{P}(X>\lambda)}{\mathbb{P}(X>\lambda_0)}d\lambda$.

• If $X$ has a density function, then the left hand side of your inequality equals $\int_{\lambda_0}^\infty xf(x)\, dx$. Will starting from this help? Dec 11, 2013 at 17:57
Let $\lambda_0=x$. By definition, the LHS is $E[X;X\gt x]$. Assume that $X=2x$ almost surely, then the LHS is $2x$ and the RHS is $\frac32x^2$ hence the inequality proposed in the post is wrong. Furthermore, if $X=y$ with $y\gt x$, $y\to x$, then the LHS converges to $x$ while every RHS of the form $$\int_x^\infty A(z)P[X\gt z]\,\mathrm dz=\int_x^y A(z)\,\mathrm dz$$ converges to $0$, hence the corresponding inequality is wrong as well.
On the other hand, $$\int_x^\infty P[X\gt z]\,\mathrm dz=E[X-x;X\gt x],$$ hence $$P[X\gt x]\,E[X\mid X\gt x]=xP[X\gt x]+\int_x^\infty P[X\gt z]\,\mathrm dz\leqslant x+\int_x^\infty P[X\gt z]\,\mathrm dz.$$