1. ## Exponential to Weibull

Problem. An exponential distribution is a member of the Gamma Family; that is, when $\displaystyle \alpha = 1$, the pdf reduces to the exponential distribution (with $\displaystyle \lambda = 1/\beta$). Let the r.v. $\displaystyle X ~$ Exponential$\displaystyle (\beta)$, then $\displaystyle Y = X^\frac{1}{\gamma}, \gamma>0$, has a Weibull $\displaystyle (\gamma, \beta)$ distribution. Show the density function of $\displaystyle Y$ is

$\displaystyle f_Y(y|\gamma, \beta) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}$.

I have in my notes that for X~Exponential, $\displaystyle f_X(x) = \lambda e^{-\lambda x}$.

Thanks!

2. pasting....

Let $\displaystyle f_X(x) = \lambda e^{-\lambda x}I(x>0)$ with $\displaystyle \lambda = 1/\beta$.

So your density is $\displaystyle f_X(x) = {1\over \beta}e^{-x/\beta}I(x>0)$

which is how I write my exponential density, but you do need the indicator function.
Otherwise you need to say x>0.

If $\displaystyle Y = X^\frac{1}{\gamma}, \gamma>0$, then

$\displaystyle f_Y(y) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}I(y>0)$.

NOW I can prove that

$\displaystyle f_Y(y) = f_X(x)\biggl|{dx\over dy}\biggr|$ but that follows from the derivative of the cdfs.

So $\displaystyle f_Y(y) = {1\over \beta}e^{-y^{\gamma}/\beta}|\gamma y^{\gamma -1}|I(y>0)$

$\displaystyle = {\gamma\over \beta} y^{\gamma -1}e^{-y^{\gamma}/\beta} I(y>0)$

since $\displaystyle y = x^{1/\gamma}$ hence $\displaystyle x = y^{\gamma}$ and $\displaystyle {dx\over dy} =\gamma y^{\gamma -1}$

3. ## my incorrect solution

This is what I did earlier in the year. Incorporating some of your notation, I think it corresponds to what you showed me above. However, I didn't get the right answer. Can you tell me what I did wrong?

Thanks.

My incorrect "solution".
Let $\displaystyle f_X(x) = \lambda e^{-\lambda x}, \forall x>0$, and $\displaystyle \lambda = \frac{1}{\beta}$.

By substituting into the Gamma density function, $\displaystyle Gamma (\alpha=1, \beta=1/\lambda)$,
we get $\displaystyle *f_X(x) = \frac{1}{\Gamma(\alpha)\beta^\alpha} x^{\alpha-1} e^{-x/\beta} = \frac{1}{\beta} e^{-x/\beta}, \forall x>0$.

So $\displaystyle F_Y(y) = P(Y \leq y)$, since $\displaystyle y = x^{1/\gamma}$, this means
$\displaystyle =P(X^{1/\gamma} \leq y)$
$\displaystyle =P(X \leq y^\gamma)$
$\displaystyle =F_X(y^\gamma)$.

Thus $\displaystyle F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1}$.

By substituting $\displaystyle y^\gamma$ for $\displaystyle x$ in $\displaystyle *f_X(x)$,
we get $\displaystyle f_Y(y) = \frac{1}{\beta} e^{-y\gamma/\beta}, \forall y>0$.

The end of my incorrect solution....

4. It's perfect except the last line.
you don't need the $\displaystyle \forall y>0$ by the way, y>0 is fine.
But you left out the chain rule which you computed.............

Thus $\displaystyle f_Y(y)=F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1}$.

By substituting $\displaystyle y^\gamma$ for $\displaystyle x$ in $\displaystyle *f_X(x)$,

AND multiplying by the $\displaystyle \gamma y^{\gamma-1}$

we get $\displaystyle f_Y(y) = \frac{1}{\beta} e^{-y^\gamma/\beta}\gamma y^{\gamma-1} , y>0$.

which is the same as mine...

This.... $\displaystyle f_Y(y)=F_Y'(y) = f_X(y^\gamma)\gamma y^{\gamma-1}$.

is $\displaystyle f_Y(y)=f_X(y^\gamma){dx\over dy}$.

5. ## last line

Would you show me how my last line should read? And why doesn't my last line look like this
$\displaystyle f_Y(y|\gamma, \beta) = \frac{\gamma}{\beta}y^{\gamma-1} e^\frac{-y^\gamma}{\beta}$?

$\displaystyle f_Y(y) = \frac{1}{\beta} e^{-y\gamma/\beta}, y>0$.