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Given normal random variable $X$ and $Y=aX+b$, the cumulative distribution function for $Y$ is $F_{Y}(y)=P(Y\leq y)$. Show that $F_{Y}(y)=P(X\leq\frac{y-b}{a})$.

I know that the cdf of the normal random variable is $\frac{1}{\sqrt{2}}\int_{-\infty}^{x}e^{-t^2/2}dt$ but I'm not sure how to go from there to the desired result.

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1 Answer 1

up vote 3 down vote accepted

There is almost nothing to do. We have $$P(Y\le y)=P(aX+b \le y)=P(aX \le y-b).$$ If $a$ is positive, then $$P(aX\le y-b)=P\left(X\le \frac{y-b}{a}\right),$$ and we are finished. Normality has nothing to do with the result.

Remark: The expression $\frac{y-b}{a}$ obviously makes no sense if $a=0$. If $a <0$, the inequality of the problem is false. In that case, since dividing by a negative reverses inequalities, we get $$P(Y\le y)=P\left(X\ge \frac{y-b}{a}\right).$$

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