If X and Y are independent random variables that are normally distributed (and therefore also jointly so), then their sum is also normally distributed.

That is $X \sim N(\mu_1,\sigma^2_1)$, $Y \sim N(\mu_2,\sigma^2_2)$ and $Z = X + Y$.

Then $Z \sim N(\mu_1+\mu_2, \sigma^2_1+\sigma^2_2)$

Is there any similar result for a positive normal distribution? That is, a normal distribution whose $f_X$ is:

\begin{cases} \frac{2}{\sqrt{2\pi}}e^{-x^2/2} & x\geq 0 \\ 0 & x < 0 \end{cases}

Does the same result apply? Thanks in advance.

  • You can try to directly compute the convolution, if you wish. But I don't think it works out. – Ian Oct 17 '16 at 17:16
up vote 0 down vote accepted

No.

To compute the pdf for the sum of two normalized Gaussians, you have

$$\int e^{-t^2/2}e^{-(x-t)^2/2}dt=\int e^{-(t^2+(x-t)^2)/2}dt=\int e^{-(t-x/2)^2}e^{-x^2/4}dt.$$

As the integral is taken over the whole of $\mathbb R$, you can freely shift the argument of the first exponential and you get a Gaussian $(0,\sqrt2)$.

The translation doesn't work when the integration range is limited to the positive reals and the antiderivative will involve the error function.

This is confirmed by Wolfram https://www.wolframalpha.com/input/?i=integrate+exp(-(t%5E2%2B(x-t)%5E2)%2F2)+for+t%3D0+to+infinity

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