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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.

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    $\begingroup$ 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, 2016 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, 2016 at 10:16

3 Answers 3

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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?

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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 $$Y\sim N(a\mu+b,a^2\sigma^2)$$

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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.

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