I have a random variable $X\sim \text{Normal}(\mu,\sigma)$ and have the transformation $Y=\max\{0,X\}$. Is the distribution of $Y$ a truncated normal where it is truncated to live on the positive half of the real line?

Secondly, is there a closed form solution for the joint distribution of a Normal random variable and a truncated Normal random variable?

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    $\begingroup$ $Y$ is a "censored" version of $X$, not a "truncated" one, to follow established terminology -because, while it is defined in the positive half of the real line (as would be $X$ truncated from below at zero), it accumulates positive probability mass at point zero. See also this post:stats.stackexchange.com/questions/77878/… $\endgroup$ Commented Jan 16, 2014 at 18:07
  • $\begingroup$ Ok, makes sense. Secondly if I have another variable $Z\sim\text{Normal}(a,b^2)$ is there a way to calculate the joint distribution of $(Y,Z)$? $\endgroup$
    – user30490
    Commented Jan 16, 2014 at 19:58
  • $\begingroup$ You can easily adapt the density found in the post I mentioned to your case, to obtain the density of $Y$. Then, if $Z$ is independent of $Y$, you just take their product. $\endgroup$ Commented Jan 16, 2014 at 20:52
  • $\begingroup$ And if they are not independent? $\endgroup$
    – user30490
    Commented Jan 16, 2014 at 21:09
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    $\begingroup$ If $X$ and $Z$ were dependent random variables then mustn't $Y$ and $Z$ be dependent random variables irrespective of whether or not $X>0$ or $X\leq 0$? $\endgroup$
    – user30490
    Commented Jan 16, 2014 at 21:32

1 Answer 1


We have to consider cases. Denote $B(y,z;\rho)$ the joint bivariate normal cumulative distribution function of two correlated variables with correlation coefficient $\rho$.

The joint support is $[0,\infty] \times (-\infty, \infty)$

For $\{Y=0, Z \in (-\infty, \infty)\}$ we have

$$P(Y=0, Z\le z) = P(X\le 0 , Z\le z) = \int_{-\infty}^{z}\int_{-\infty}^0f_{XZ}(x,z)dxdz$$

$$= B(0,z;\rho_{XZ})$$

For $\{Y>0, Z \in (-\infty, \infty)\}$ we have $$P(Y>0, Z\le z) = P(0<X\le x , Z\le z) = \int_{-\infty}^{z}\int_{0}^xf_{XZ}(x,z)dxdz$$ $$= B(x,z;\rho_{XZ}) = B(y^+,z;\rho_{XZ})$$

the last equality because for this range $Y = X$.

Bringin together using indicator functions

$$F_{YZ}(y,z) = B(0,z;\rho_{XZ})\cdot I_{\{Y=0\}} + B(y^+,z;\rho_{XZ})\cdot (1-I_{\{Y=0\}})$$

Differentiation of the two branches of the cdf will give you the corresponding densities.

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    $\begingroup$ Strictly speaking, this CDF doesn't have a density because it is singular along a line $Y=0$. $\endgroup$
    – whuber
    Commented Jan 16, 2014 at 23:44
  • $\begingroup$ What should I do @Whuber? $\endgroup$
    – user30490
    Commented Jan 16, 2014 at 23:46
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    $\begingroup$ Do you need to do anything? Alecos has given you an explicit solution in terms of standard mathematical objects; that's a reasonable interpretation of "closed form" and it allows for simple calculation of the probability of any event. $\endgroup$
    – whuber
    Commented Jan 16, 2014 at 23:59
  • $\begingroup$ @Whuber, is his solution correct? $\endgroup$
    – user30490
    Commented Jan 17, 2014 at 0:03
  • $\begingroup$ Also, are (Y,Z) jointly bivariate Normal as he states above? $\endgroup$
    – user30490
    Commented Jan 17, 2014 at 0:04

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