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{aX+b≤y}=Pr{X≤(y−b)/a}=∫(y−b)/a−∞fX(x)dx.
Using the substitution x=(t−b)/a, we obtain that
Pr{Y≤y}=∫(y−b)/a−∞fX(x)dx=∫y−∞1afX((t−b)/a)dt
and the density function fY(y)=a−1fX((y−a)/b).
In general, if Y=g(X) with a monotone function g, we have that
fY(y)=|ddy(g−1(y))|⋅fX(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∈R.
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