# Cumulative distribution function question [duplicate]

Possible Duplicate:
Normal random variable $X$ and the cdf of $Y=aX+b$

I'm given a standard random variable $X$, and $Y = aX + b$:

How can I find the cumulative distribution function for Y as an integral of $f(x)=(\frac{1}{\sqrt{2}\pi\sigma}e^{-\frac{(x-u)^2}{\sigma^2}}$?

I know $F_y(y)=F_x(\frac{y-b}{a})$, but cant figure out where to go from there.

-
You should write $F_Y(y)= F_X((y-b)/a)$, with capital letters in the subscripts referring to the random variables involved, and lower-case letters as the arguments to the functions. That way you're not using the same letter for two different things. This is important when you want to refer to something like $\Pr(Y \le y)$. –  Michael Hardy Feb 16 '12 at 13:43
Are you sure your expression for $F_Y(y)$ is correct in the case $a < 0$ also? But, assuming $a > 0$, does it not follow from the definition of $F_X(t)$: $$F_X(t) = \int_{-\infty}^t f_X(x) \mathrm dx$$ that $$F_Y(y) = F_X\left(\frac{y-b}{a}\right) = \int_{-\infty}^{\frac{y-b}{a}} f_X(x) \mathrm dx ?$$ Straight plug-and-chug, no mess, and it works even when $X$ is not a normal random variable. By the way, your density for $X$ is not that of a standard normal random variable. –  Dilip Sarwate Feb 16 '12 at 14:58
See also the answer to this almost identical question posted about an hour before yours. Common homework? –  Dilip Sarwate Feb 16 '12 at 17:07

## marked as duplicate by Qiaochu YuanFeb 16 '12 at 18:04

As an integral of the function you have given, you would need $u=b$ and $\sigma=a$.

Proof: As you note

$$F_Y(y) = F_X\left(\frac{y-b}{a}\right) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{t} e^{-x^2/2} \,dx,$$

where $t = \frac{y-b}{a}$. Now make the substitution $x = \frac{s-b}{a}$. Then $dx = ds/a$, so

$$F_Y(y) = \frac{1}{a\sqrt{2\pi}}\int_{-\infty}^{y} e^{-(s-b)^2/2a^2} \,ds.$$

-
Be careful when $a < 0$; you don't want a random variable whose standard deviation $\sigma$ is negative. –  Dilip Sarwate Feb 16 '12 at 15:58
If $a>0$ then \begin{align} & f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy}\Pr(Y\le y) = \frac{d}{dy}\Pr(aX+b\le y) = \frac{d}{dy}\Pr\left(X \le \frac{y-b}{a}\right) \\ \\ & = \frac{d}{dy}F_X\left( \frac{y-b}{a} \right) = f_X\left(\frac{y-b}{a}\right) \cdot \frac{d}{dy} \frac{y-b}{a} = \frac 1 a f_X\left(\frac{y-b}{a}\right) \\ \\ \\ & = \frac 1 a \cdot \frac{1}{\sqrt{2\pi}}\exp\left(\frac{-1}{2} \cdot \left(\frac{y-b}{a}\right)^2\right). \end{align}
Therefore $$F_Y(y) = \frac 1 a \int_{-\infty}^y \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-1}{2} \left(\frac{u-a}{a}\right)^2\right)\; du.$$
The equivalence between $aX+b\le y$ and $X \le \frac{y-b}{a}$ holds only if $a>0$.