1.let $X$ be a random variable , $A\subset B$ is that $\mathbb{E} [X|A] \leq \mathbb{E}[X|B]$ true in general ?

2 what about If $X$ is Gaussian random variable , $A\subset B$ is that $\mathbb{E} [X|X \leq 0] \leq \mathbb{E}[X| X\leq c]$ where $c>0$?


The first point is not true.

Consider $X$ being a uniform rv on $\{0,1\}$, then $$1=E(X|\{X \in \{1\}\})>E(X|\{X \in \{0,1\}\})=\frac 1 2.$$

The second one is true for $c \ge 0$ since $$\mathbb{E} [X|X \leq 0] = \int_{-\infty}^0 xf(x)dx \leq \int_{-\infty}^cxf(x)dx=\mathbb{E}[X| X\leq c]$$

This follows from the fact that : $xf(x)>0$ on $(0,c)$.


Just add as a slight generalization of the problem:

Assume $\mathbb{P}\{0 < X \leq c\} > 0$. Consider

$$ \begin{align} \mathbb{E}[X \mid X \leq c] &= \mathbb{E}[X \mid X \leq 0, X\leq c]\mathbb{P}\{X \leq 0 \mid X \leq c\} \\ &~~~~+ \mathbb{E}[X \mid 0 < X \leq c, X\leq c]\mathbb{P}\{0 < X \leq c \mid X \leq c\} \\ & = \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\}} {\mathbb{P}\{X \leq c\}} + \mathbb{E}[X \mid 0 < X \leq c] \frac {\mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & > \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\}} {\mathbb{P}\{X \leq c\}} + \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & = \mathbb{E}[X \mid X \leq 0] \frac {\mathbb{P}\{X \leq 0\} + \mathbb{P}\{0 < X \leq c\}} {\mathbb{P}\{X \leq c\}} \\ & = \mathbb{E}[X \mid X \leq 0] \end{align}$$

The first line is due to the law of probability.

Since $\mathbb{P}\{X \leq 0 \mid X \leq 0\} = 1$ and $\mathbb{P}\{X > 0 \mid 0 < X \leq c \} = 1$, we have $$\mathbb{E}[X \mid X \leq 0] \leq 0 < \mathbb{E}[X \mid 0 < X \leq c]$$ and this results in the inequality in the 3rd line.

The constant $0$ here is relatively arbitrary, and in general we have $$ \mathbb{E}[X \mid X \leq b] \leq \mathbb{E}[X \mid X \leq c]$$ whenever $b < c$, and $\mathbb{P}\{b < X \leq c\} > 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.