Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X_1, \dots, X_n$ are truncated standard normal variables, truncated so that $X_i \geq 0$ (that is, $X_i$ is drawn as a standard normal, conditional on $X_i \geq 0$)

Let $c_1, \dots, c_n$ be non-negative coefficients.

What does the distribution of $\sum_i c_i Y_i$ look like? Does it have, or approximately have, a standard distribution, such as a truncated normal distribution?

Original question:
Suppose $X_1, \dots, X_n$ are iid Normal random variables, with mean 0 and variances $\sigma_1, \dots, \sigma_n$.

Let $Y_i = \max(0,X_i)$. (So $Y_i$ is a truncated normal random variable).

What does the distribution of $\sum_i Y_i$ look like? Does it have, or approximately have, a standard distribution?

share|cite|improve this question
You edited the question, based on my answer, in such a way that makes my answer look irrelevant. This is a bad practice, because it wasted my effort. The accepted way was to leave the original question in place, probably at the bottom of the question. – Sasha Jul 25 '12 at 13:12

This post answers the original question.

Please be careful, $Y_k$ is not a truncated normal random variable, it is censored normal random variable. In particular: $$ \mathbb{P}(Y_k = 0) = \mathbb{P}\left(X_k \leqslant 0 \right) = \frac{1}{2} \not= 0 $$ meaning that $Y_k$ is not an absolutely continuous random variable. Rather, $Y_k$ can be thought of as the mixture of a degenerate random variable, concentrated at $x=0$, and a normal random variable, truncated to the positive semi-axis.

With this said, $Z = \sum\limits_{k=1}^n Y_k$ is not absolutely continuous either, since $$ \mathbb{P}\left(Z=0\right) = \mathbb{P}\left(X_1 \leqslant 0, \ldots, X_n \leqslant 0\right) = \mathbb{P}\left(X_1 \leqslant 0\right) \cdots \mathbb{P}\left(X_n \leqslant 0\right) = \frac{1}{2^n} $$ The absolutely continuous part of $Z$ is not equal in distribution to any standard distribution. One can compute the characteristic function of $Z$ rather easily. $$ \begin{eqnarray} \phi_{Y_k}(t) &=& \mathbb{E}\left(\exp\left(i \max(0,X_k)t\right)\right) = \mathbb{P}\left(X_k \leqslant 0\right) + \mathbb{E}\left(\exp\left(i X_k t\right): X_k > 0\right)\\ &=& \frac{1}{2} + \exp\left(-\frac{t^2}{2} \sigma_k^2\right) + \frac{2 i}{\sqrt{\pi}} D_F\left( \frac{t \sigma_k}{\sqrt{2}} \right) \end{eqnarray}$$ where $D_F(x)$ denotes Dawson's F-function.

Since the random variables $Y_k$ are independent: $$ \phi_Z(t) = \phi_{Y_1}(t) \cdots \phi_{Y_n}(t) $$

Here is a histogram for several small values of $n$, assuming equal unit variance: enter image description here

You can explicitly see how the cumulative distribution function is not continuous at $x=0$, and how the size of the discontinuity jump decreases as $n$ grows.

share|cite|improve this answer
+1. Replaced some indexes $i$ by $k$ to avoid the confusion with the complex number $i$. (But you might want to check the formula for $\phi_{Y_k}$ which involves $D_F$.) – Did Jul 25 '12 at 14:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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