If $X, Y \sim N(0,1)$, find the CDF of $\alpha X + \beta Y$ 
Possible Duplicate:
Proof that the sum of two Gaussian variables is another Gaussian 

Let $X,Y$ be independent normally distributed $N(0,1)$ random variable, and $\alpha,\beta\in \mathbb{R}$. What is the cumulative distribution function of $\alpha X+\beta Y$? 
Thank you very much for your help. 
 A: We have
$$\operatorname{var}(\alpha X + \beta Y) = \alpha^2 \operatorname{var}(X) + \beta^2\operatorname{var}(Y).$$
(If they were correlated, we'd have needed a third term: $\operatorname{cov}(\alpha X,\beta Y) = \alpha\beta\operatorname{cov}(X,Y)$.)
So the standard deviation of the distribution you're looking for is $\sqrt{\alpha^2+\beta^2}$.  And the expected value is $0$.
So we get
$$
F_{\alpha X + \beta Y}(x) = \Pr(\alpha X + \beta Y\le x) = \Pr\left(\frac{\alpha X + \beta Y}{\sqrt{\alpha^2+\beta^2}} \le \frac{x}{\sqrt{\alpha^2+\beta^2}}\right) = \Pr\left(Z \le \frac{x}{\sqrt{\alpha^2+\beta^2}}\right)
$$
where, as usual, $Z\sim N(0,1)$.  This may be written as
$$
\Phi\left(\frac{x}{\sqrt{\alpha^2+\beta^2}}\right)
$$
where, as usual, $\Phi$ is the standard normal c.d.f.
A: It looks from your comment as if the meaning of your question is different from what I thought at first.  My first answer assumed you knew that the sum of independent normals is itself normal.
You have
$$
\exp\left(-\frac12 \left(\frac{x}{\alpha}\right)^2 \right) \exp\left(-\frac12 \left(\frac{z-x}{\beta}\right)^2 \right)
= \exp\left(-\frac12 \left( \frac{\beta^2x^2 + \alpha^2(z-x)^2}{\alpha^2\beta^2} \right) \right).
$$
Then the numerator is
$$
\begin{align}
& (\alpha^2+\beta^2)x^2 - 2\alpha^2 xz + \alpha^2 z^2 \\  \\
& = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz\right) + \alpha^2 z^2 \\  \\
& = (\alpha^2+\beta^2)\left(x^2 - 2\frac{\alpha^2}{\alpha^2+\beta^2} xz + \frac{\alpha^4}{(\alpha^2+\beta^2)^2}z^2\right) + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2 \\  \\
& = (\alpha^2+\beta^2)\left(x - \frac{\alpha^2}{\alpha^2+\beta^2}z\right)^2 + \alpha^2 z^2 - \frac{\alpha^4}{\alpha^2+\beta^2}z^2,
\end{align}
$$
and then remember that you still have the $-1/2$ and the $\alpha^2\beta^2$ in the denominator, all inside the "exp" function.
(What was done above is completing the square.)
The factor of $\exp\left(\text{a function of }z\right)$ does not depend on $x$ and so is a "constant" that can be pulled out of the integral.
The remaining integral does not depend on "$z$" for a reason we will see below, and thus becomes part of the normalizing constant.
If $f$ is any probability density function, then
$$
\int_{-\infty}^\infty f(x - \text{something}) \; dx
$$
does not depend on "something", because one may write $u=x-\text{something}$ and then $du=dx$, and the bounds of integration are still $-\infty$ and $+\infty$, so the integral is equal to $1$.
Now look at
$$
\alpha^2z^2 - \frac{\alpha^4}{\alpha^2+\beta^2} z^2 = \frac{z^2}{\frac{1}{\beta^2} + \frac{1}{\alpha^2}}.
$$
This was to be divided by $\alpha^2\beta^2$, yielding
$$
\frac{z^2}{\alpha^2+\beta^2}=\left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2.
$$
So the density is
$$
(\text{constant})\cdot \exp\left( -\frac12 \left(\frac{z}{\sqrt{\alpha^2+\beta^2}}\right)^2 \right) .
$$
Where the standard deviation belongs we now have $\sqrt{\alpha^2+\beta^2}$.
