What is the difference between $X^+$ and $X | X > 0$ where $X \sim N(0,1)$

Question

I am confused between $X^+$ and $X | X > 0$ where $X \sim N(0,1)$.

Here we see the usage of $X^+$

where $X^+ \max(0, X)$. And we see that $E[X^+] = \int_0^{\infty}x N(0,1)dx = 1/\sqrt{2\pi}$

Here

we see $E[X | X > 0] = \int_0^{\infty} x * 2 \cdot N(0,1)dx = \frac{\sqrt{2}}{\sqrt{\pi}}$ where $f_{X | X > 0} = 2 \cdot N(0,1)$.

We also can find that the latter case $E[X | X > 0] = E[|X|]$.

But I am confused here by what is the difference between $X^+$ and $X | X > 0$? They seem to be the same thing, but obviously their expectations are different by a factor of 2 (assuming they're computed correctly)

Answer 1

$$X^+ = \begin{cases} X, & X > 0 \\ 0, & X \le 0. \end{cases}$$ Whereas, $$X \mid X > 0 = \begin{cases} X, & X > 0 \\ \text{undefined}, & X \le 0. \end{cases}$$ If we let $Y_1 = X^+$ and $Y_2 = X \mid X > 0$, then the support of $Y_1$ is $\mathbb R$, but the support of $Y_2$ is $\mathbb R^+$. This means there is a discrete probability mass for $Y_1$, specifically $\Pr[Y_1 = 0] = 1/2$. $Y_1$ can be thought of as a left-censored variable: values of $X$ less than or equal to $0$ are observed as $0$. $Y_2$, on the other hand is a left-truncated variable: values of $X$ less than or equal to $0$ are never observed at all.

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A simple example may be in order.

Suppose I have a fair six-sided die with faces numbered from $1$ to $6$ inclusive. I roll the die but you can't see it. All you know is what I tell you about the outcome.

In the first scenario, I tell you in advance that if I roll a number less than $3$, I will report the result as $3$. What is the resulting probability distribution that you observe?

In the second scenario, I tell you in advance that if I roll a number less than $3$, I do not report the result, but instead re-roll the die until I obtain a number that is at least $3$, at which point I report that value. You are never told whether I re-rolled. What is the resulting distribution of outcomes you observe?

In both scenarios, I have access to the complete information about the outcomes, so what I observe is discrete uniform on $\{1, \ldots, 6\}$. But what you observe in the first scenario is not this. Instead, the outcome is $3$ with probability $1/2$ and $4$, $5$, or $6$ each with probability $1/6$. In the second scenario, you observe a discrete uniform distribution on $\{3, 4, 5, 6\}$, each with probability $1/4$.

Answer 2

By definition $E[X|A]=\frac {E[XI_A]} {P(A)}$. So $E[X|X>0]$ is nothing but $EX^{+}$ divided by $P(A)$ where $A=(X>0)$. Note that $P(A)=\frac 1 2$.