Convolutions and the Gaussian distribution Suppose $X_1$ and $X_2$ are independent random variables each with the standard Gaussian distribution. Compute, using convolutions, the density of the distribution of $X_1 + X_2$ and show $X_1 + X_2 = \sqrt{2}X$ where X has standard Gaussian distribution.
I wrote down the density of the distribution as $f_{x_1+x_2}(z)= \int_{-\infty}^{\infty}f_{x_1}(x)f_{x_2}(z-x)dx$, and I don't know how to proceed to the next part of the question.
Any help is appreciated, thanks in advance.
 A: For $X, Y\sim N(0,1)$, we have
$${f_{X + Y}}\left( x \right) = \int_{ - \infty }^\infty  {{f_X}\left( {x - y} \right){f_Y}\left( y \right)dy}  = \frac{1}{{2\pi }}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - \frac{1}{2}{{\left( {x - y} \right)}^2}}}{e^{ - \frac{1}
{2}{y^2}}}dy}  = \frac{1}{{2\pi }}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {y^2}} \right]}}dy}.$$
We have
$${\left( {x - y} \right)^2} + {y^2} = {x^2} - 2xy + 2{y^2} = {\left( {\sqrt 2 y - \frac{1}{{\sqrt 2 }}x} \right)^2} + \frac{{{x^2}}}{2}.$$
So
$$\begin{gathered}
  {f_{X + Y}}\left( x \right) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - \frac{1}{2}\left[ {{{\left( {\sqrt 2 y - \frac{1}
{{\sqrt 2 }}x} \right)}^2} + \frac{{{x^2}}}{2}} \right]}}dy}  = \frac{1}
{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - \frac{1}{2}{{\left( {\sqrt 2 y - \frac{1}{{\sqrt 2 }}x} \right)}^2}}}dy}  \hfill \\
   = \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - {{\left( {\frac{{\sqrt 2 y - \frac{1}{{\sqrt 2 }}x}}
{{\sqrt 2 }}} \right)}^2}}}dy}  = \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}
{4}}}\int_{ - \infty }^\infty  {{\operatorname{e} ^{ - {{\left( {y - \frac{1}
{2}x} \right)}^2}}}d\left( {y - \frac{1}{2}x} \right)}  \hfill \\
   = \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\sqrt \pi   = \frac{1}{{2\sqrt \pi  }}{e^{ - \frac{{{x^2}}}{4}}}. \hfill \\ 
\end{gathered}$$
Now, for $Z\sim N(0,1)$, wet have
$${F_{\sqrt 2 Z}}\left( x \right) = P\left( {\sqrt 2 Z < x} \right) = P\left( {Z < \frac{x}
{{\sqrt 2 }}} \right) = {F_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right)$$
This leads to
$$\begin{gathered}
  {f_{\sqrt 2 Z}}\left( x \right) = \frac{d}
{{dx}}\left[ {{F_{\sqrt 2 Z}}\left( x \right)} \right] = \frac{d}
{{dx}}\left[ {{F_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right)} \right] = \frac{1}
{{\sqrt 2 }}{{F'}_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right) = \frac{1}
{{\sqrt 2 }}{f_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right) \hfill \\
   = \frac{1}
{{\sqrt 2 }}.\frac{1}
{{\sqrt {2\pi } }}{e^{ - \frac{1}
{2}{{\left( {\frac{x}
{{\sqrt 2 }}} \right)}^2}}} = \frac{1}
{{2\sqrt \pi  }}{e^{ - \frac{{{x^2}}}
{4}}} \hfill \\ 
\end{gathered} $$
We deduce that ${f_{\sqrt 2 Z}}\left( x \right) = {f_{X + Y}}\left( x \right)$.
A: $$\begin{eqnarray*}\frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-x^2/2}e^{-(z-x)^2/2}\,dx &=& \frac{1}{2\pi}\int_{-\infty}^{+\infty}\exp\left(-\frac{2x^2-2xz+z^2}{2}\right)\,dx\\&=&\frac{1}{2\pi}\int_{-\infty}^{+\infty}\exp\left(-x^2-\frac{z^2}{4}\right)\,dx\\&=&\frac{1}{2\sqrt{\pi}}\,e^{-z^2/4}. \end{eqnarray*}$$
