Results 1 to 6 of 6

Thread: Exponential to Weibull

  1. #1
    Member
    Joined
    Mar 2008
    From
    Texas
    Posts
    110
    Awards
    1

    Question Exponential to Weibull

    Can you please help?

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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 $\displaystyle {dx\over dy} =\gamma y^{\gamma -1}$
    Last edited by matheagle; Dec 13th 2009 at 07:01 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    From
    Texas
    Posts
    110
    Awards
    1

    Arrow 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....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    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 $\displaystyle f_Y(y)=f_X(y^\gamma){dx\over dy}$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    From
    Texas
    Posts
    110
    Awards
    1

    last line

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

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

    (LaTex ?: how do I add a space after the "," and "y" in my last line above?)

    Thanks for everything!!!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor matheagle's Avatar
    Joined
    Feb 2009
    Posts
    2,763
    Thanks
    5
    GN poisson ivy
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. CDF of the Weibull Distribution
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Apr 17th 2013, 02:06 AM
  2. Weibull or Poisson?
    Posted in the Advanced Statistics Forum
    Replies: 12
    Last Post: Mar 4th 2012, 11:06 AM
  3. Weibull & Exponential distribution
    Posted in the Statistics Forum
    Replies: 3
    Last Post: Jun 19th 2010, 04:14 AM
  4. Show X~Exponential has a Weibull distribution.
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Oct 20th 2009, 02:36 PM
  5. weibull parameter
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: Oct 30th 2008, 01:19 PM

Search Tags


/mathhelpforum @mathhelpforum