3
$\begingroup$

Let $X$ be continuous random variable - $X \sim N(\mu,\sigma^2)$.

Let $Y$ be continuous random variable when $Y = a X + b$.

Prove that $Y$ distributes normal distribution also, and find $\mu_Y,\sigma_Y^2$.

My Attempt-

$F_Y(y) = P(Y \leq y) = P(aX+b \leq y) = P(X\leq \frac{y-b}{a}) = F_X(\frac{y-b}{a}).$

And then- $f_Y = \frac{dF_Y(y)}{dy} = \frac{dF_X(\frac{y-b}{a})}{dy} = \frac{f_X(\frac{y-b}{a})}{\frac{y}{a}}$.

But, I don't know how to continue. Can someone help me? Thanks.

$\endgroup$
  • 1
    $\begingroup$ You made a mistake when computing $f_Y$. It should be $f_Y(y) = \frac{f_X(\frac{y-b}{a})}{a}$ Then try to replace $f_X$ by its actual definition. $\endgroup$ – Dark Jul 6 '16 at 10:08
  • $\begingroup$ Actually an extra condition is needed: $a\neq0$. Expectation and variance of $Y$ can be found directly in the sense that "normal distribution" is irrelevant. $\endgroup$ – drhab Jul 6 '16 at 10:16
1
$\begingroup$

Other way $$M_{Y}(t)=\mathbb{E}[e^{tY}]=\mathbb{E}[e^{t(aX+b)}]=e^{bt}\mathbb{E}[e^{atX}]=\large e^{bt}\, e^{a\mu\,t+\frac12\sigma^2a^2t^2}=e^{(a\mu+b)t+\frac12(a\sigma)^2t^2}$$ therefore $$Y\sim N(a\mu+b,a^2\sigma^2)$$

$\endgroup$
4
$\begingroup$

Given $X \sim N(\mu, \sigma^2)$ then as you have derived, one obtains $$F_y(y) = F_X\left(\frac{y-b}{a}\right)$$ Therefore, the density of $Y$ is given by \begin{align} f_Y(y) &= \frac{1}{a} f_X\left(\frac{y-b}{a}\right) \\ &= \frac{1}{\sqrt{2\pi a^2 \sigma^2}} \exp\left[\frac{-1}{2\sigma^2} \left(\frac{y-b}{a}-\mu\right)^2 \right] \\ &= \ldots \end{align} Can you proceed?

$\endgroup$
1
$\begingroup$

Start with $U\sim Norm\left(0,1\right)$ and let $\sigma>0$.

If $X=\sigma U+\mu$ then we find $F_{X}\left(x\right)=P\left(U\leq\frac{x-\mu}{\sigma}\right)=\Phi\left(\frac{x-\mu}{\sigma}\right)$ and $f_{X}\left(x\right)=\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)$ wich can be recognized as the PDF connected with distribution $Norm\left(\mu,\sigma^2\right)$.

If $a>0$ and $Y=aX+b$ then $Y=a\sigma U+\left(a\mu+b\right)=\sigma'U+\mu'$ and the same procedure can be applied.

If $a<0$ and $Y=aX+b$ then $Y=\left(-a\right)\sigma V+\left(a\mu+b\right)=\sigma''V+\mu'$ where $V=-U$. Then $V\sim Norm\left(0,1\right)$ and the same procedure can be applied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.