Let X be a continuous random variable with density function fX and let a,b>0.
What is Y=aX+b?
I need some help with this one. And I am quite sure it is not afX+b.
Answer
We have that
Pr
Using the substitution x=(t-b)/a, we obtain that
\Pr\{Y\le y\}=\int_{-\infty}^{(y-b)/a}f_X(x)\mathrm dx=\int_{-\infty}^y\frac1af_X((t-b)/a)\mathrm dt
and the density function f_Y(y)=a^{-1}f_X((y-a)/b).
In general, if Y=g(X) with a monotone function g, we have that
f_Y(y) = \left| \frac{\mathrm d}{\mathrm dy} (g^{-1}(y)) \right| \cdot f_X(g^{-1}(y)),
where g^{-1} denotes the inverse function (see here for more details). In this particular case g(x)=ax+b for x\in\mathbb R.
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