1. ## Exponential to Weibull

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

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

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

Thanks!

2. pasting....

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

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

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

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

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

NOW I can prove that

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

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

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

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

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 $f_X(x) = \lambda e^{-\lambda x}, \forall x>0$, and $\lambda = \frac{1}{\beta}$.

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

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

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

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

The end of my incorrect solution....

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

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

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

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

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

which is the same as mine...

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

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

5. ## last line

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

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