Find $\mathbb E[X|X>0]$ of a symmetric continuous distribution Find $\mathbb E[X|X>0]$ of a symmetric continuous distribution $(-\infty,\infty)$ , where $\mathbb E[X]=0$ and $\mathsf{Var}[X]=\sigma^2$
I tried to split the integral at $0$ and so the integral from $0$ to $\infty$, and so I came up with a value of $0.5....$
It doesn't seem to work too well... 
 A: Consider the four-point distribution $\pm\varepsilon,\pm M$, where $\varepsilon\ll\sigma\ll M$.  With appropriate choice of probability nasses we can achieve zero-mean and variance $\sigma^2$, and any $\mathbb{E}|X|\in(0,\sigma)$.
Now smooth things out a little, say little triangles on $[\frac12\varepsilon,\frac32\varepsilon]$ and $[\frac12M,\frac32M]$ (and by symmetry to the negative), and choose probablity weights approriately gives any $\mathbb{E}[X\mid X>0]\in(0,\sigma)$.
To show you can't get anything $\geq\sigma$, use Cauchy-Schwarz.
A: Let $A$ be the event $X>0$
Then, by Jensen inequality (or by non-negativity of conditioned variance): $E[X^2 \mid A] \ge (E[X\mid A])^2 $
Because of symmetry, $E[X^2 \mid A]=E[X^2]=\sigma^2 $.
Hence $E[X\mid A] \le \sigma$
You cannot expect to find more than a bound here. 
A: If $f(x)$ is the symmetric and continuous pdf of $X$, then:
$$g(x) = \begin{cases}
2f(x) & \text{if}~x > 0\\
0 & \text{else}
\end{cases},$$
is the conditional distribution of $X$ knowing that $X > 0$.
Clearly:
$$a = \mathbb{E}[X | X>0] = \int_{-\infty}^{+\infty} xg(x)dx = 2\int_{0}^{+\infty} xf(x)dx > 0,$$
and
$$b = \mathbb{E}[X^2 | X>0] = \int_{-\infty}^{+\infty} x^2g(x)dx = 2\int_{0}^{+\infty} x^2f(x)dx > 0.$$
Now, observe that $x^2f(x)$ is an even function. Therefore:
$$\sigma^2 = \int_{-\infty}^{+\infty} x^2f(x)dx = 2\int_{0}^{+\infty} x^2f(x)dx = b.$$
The variance of $X$ knowing that $X > 0$ is:
$$\mathbb{E}[X^2 | X>0] - \mathbb{E}[X | X>0]^2 = b-a^2 = \sigma^2 - a^2 \geq 0.$$
This means that:
$$a^2 \leq \sigma^2 \Rightarrow 0 < a \leq \sigma,$$
or equivalently:
$$0 < \mathbb{E}[X | X>0] \leq \sigma.$$ 
