# Find the density function of $Z=X+Y$ when $X$ and $Y$ are standard normal [duplicate]

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Two independent random variables $X$ and $Y$ have standard normal distributions. Find the density function of random variable $Z=X+Y$ using the convolution method.

This shows this problem on page 4 but it seems to use some tricks that are, well, pretty tricky. In particular getting behind their third line is beyond me. Is there an alternate way to show this that is more straightforward perhaps even with more (unskipped) steps?

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## 2 Answers

$$f_{Z}(z) = \int_{-\infty}^{\infty} f_X(z - y)f_Y(y)dy \\= \frac{1}{2 \pi}\int_{-\infty}^{\infty} e^{-\frac{(z - y)^2}{2}}e^{-\frac{y^2}{2}}dy \\= \frac{1}{2 \pi}\int_{-\infty}^{\infty} exp\left(-\frac{z^2 - 2zy + 2y^2}{2}\right)dy \\= \frac{1}{2 \pi}\int_{-\infty}^{\infty} exp\left(-\frac{\frac{z^2}{2} - 2zy + 2y^2 + \frac{z^2}{2}}{2}\right)dy \\= \frac{1}{2 \pi}\int_{-\infty}^{\infty} exp\left(-\frac{(\frac{z}{\sqrt{2}} - \sqrt{2}y)^2 + \frac{z^2}{2}}{2}\right)dy \\= \frac{e^{-\frac{z^2}{4}}}{2 \pi}\int_{-\infty}^{\infty} exp\left(-(\frac{z}{\sqrt{2}} - \sqrt{2}y)^2 / 2\right)dy \\= \frac{e^{-\frac{z^2}{4}}}{2\pi} \times \sqrt{\pi} \\= \frac{e^{\frac{-z^2}{4}}}{\sqrt{4 \pi}}$$ So, $Z \sim N(0, 2)$

A very simple way of finding the density of Z is to recognize that the sum of Gaussians is still Gaussian. Then you can identify the distribution by finding the mean and variance of Z. $${\mu _Z} = {E_Z}\left[ Z \right] = {E_X}\left[ X \right] + {E_Y}\left[ Y \right] = {\mu _X} + {\mu _Y} = 0$$ and $$\begin{array}{l} \sigma _Z^2 = {E_Z}\left[ {{Z^2}} \right] - \mu _Z^2 = {E_X}\left[ {{X^2}} \right] + 2{E_X}\left[ X \right]{E_Y}\left[ Y \right] + {E_Y}\left[ {{Y^2}} \right] - \mu _Z^2\\ = \left( {\sigma _X^2 + \mu _X^2} \right) + 2{\mu _X}{\mu _Y} + \left( {\sigma _Y^2 + \mu _Y^2} \right) - \left( {\mu _X^2 + 2{\mu _X}{\mu _Y} + \mu _Y^2} \right) = \sigma _X^2 + \sigma _Y^2 = 2 \end{array}$$