# If $X, Y \sim N(0,1)$, find the CDF of $\alpha X + \beta Y$ [duplicate]

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.

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## marked as duplicate by Did, Srivatsan, Qiaochu YuanSep 19 '11 at 2:38

This question was marked as an exact duplicate of an existing question.

so I have like $\int_{\infty} ^{\infty} \frac{1}{2\pi} \frac{1}{\alpha\beta} e^\frac{(-x/\alpha)^2}{\alpha} e^\frac{-((z-x)/\alpha)^2}{\alpha}dx$ but i have no idea how to calculate it – lovelyaagirl Sep 18 '11 at 20:32
The integral can be reduced via algebra to a well-known integral. You need to know when and how to complete a square. – Michael Hardy Sep 18 '11 at 20:41
This must be a duplicate... – cardinal Sep 18 '11 at 20:44
That page lacks a proper proof and the only answer points to the wikipedia page for a proof. Perhaps someone should post another more descriptive answer there? Or, is the wikipedia link enough? (The article seems pretty well-written and complete.) – Srivatsan Sep 18 '11 at 21:17

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.

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It is not clear (to me) what you chose to prove and what you chose to assume. Apparently you already know the result will be gaussian and you identify its mean and variance? Then, why bother including the displayed sequence of equalities between CDFs... – Did Sep 18 '11 at 20:54

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 $\sqrt{\alpha^2+\beta^2}$.

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Hummm... sorry but you should definitely check your computations. – Did Sep 18 '11 at 21:07
isnt there a z missing from the second term in the bracket in line2? – lovelyaagirl Sep 18 '11 at 21:20
Haste makes waste: there was indeed a missing $z$. To be continued....... – Michael Hardy Sep 19 '11 at 0:47
OK, now what is your opinion? – Michael Hardy Sep 19 '11 at 0:58