Expectation of non-negative random variable vs Truncated random variable

Question

I want to show that E[X|X>t]>=E[X],

where X is discrete non-negative r.v, and t is positive deterministic value.

Some ideas?

Edit: >= instead of >

Answer 1

Hints on a counter-example to $E[X|X>t]>E[X]$

Let $t=1$. Try to find a random variable $X$ for which the information $\{X>1\}$ does not tell us anything new.

Hints on proving $E[X|X>t]\geq E[X]$

We don't care what kind of random variable $X$ is. We only require that the expectations are well defined, we don't care whether or not $t$ or $X$ can take negative values.

1) Try proving that for any probability $p \in [0,1]$ we have: $$ 5p + 7(1-p)\leq 7$$

2) Try proving that for any real numbers $a,b$ that satisfy $a\leq b$, and any probability $p\in[0,1]$, we have: $$ap + b(1-p) \leq b$$

3) Try proving that $E[X|X>t]\geq t$. Try proving a related inequality for $E[X|X\leq t]$.

Answer 2

This is, I suppose, implicit in the hint sequence presented in Michael's answer.

All bets are off if $P(X>t)=0$.

Otherwise write $$EX=E[X|X>t]P(X>t) + E[X|X\le t]P(X\le t).$$ Thus the number $EX$ is an element of the closed convex hull of the two conditional expectations, that is of the interval with those conditional expectations as endpoints. It is easy to check that $E[X|X\le t]\le t \le E[X|X>t]$ and hence to see which is the right-hand endpoint.