Suppose that X ∼ Exp(1). For α > 0, β > 0, and −∞ < ν < ∞,
find
a) the pdf of $Y = αX^
{1/β} + ν$.
b) E[Y ]
Working on this problem right now but it seems complicated so I'm anticipating needing help. Will edit work into this part as I go along.
a)
$F_Y(y) = P(Y < y) = P(αx^
{1/
β} + ν < y) = P(x < ({y- ν}/α)^β) = F_X(((y- ν)/α)^β) =
\int_0^{((y- ν)/α)^β}e^{-x}dx $ -> integrate & differentiate -> weibull distribution pdf for y > v
b) Thanks for the help below!
$E[Y] = E[αX^{1/β} + ν] = αE[X^{1/β}] + ν = α\int_0^\infty x^{1/β}e^{-x}dx + ν = α\int_0^\infty x^{(1/β + 1) – 1}e^{-x}dx + ν = α\Gamma((1/ \beta) + 1) + ν = (\alpha / \beta) \Gamma(1/\beta) + ν$
Best Answer
This theorem will be useful : Let X be a continuous type random variable with pdf f(x). If h is strictly monotonic function and differentiable then the pdf of Y=h(X) is given by
$$f_Y(y) = f_X(h^{-1}(y))*\lvert dh^{-1}(y)/dy\rvert$$ if y=h(x) else 0 if y is not equal to h(x)
In this problem Y=h(X)=αX1/β+ν. h is strictly monotonic and differentiable. So by applying the theorem we get the answer to the first part as $$f_Y(y)=e^{{-((Y-v)}/α)^β}*β/α*((Y-v)/α)^{β-1}$$
and for the second part use by parts theorem.