Results 1 to 9 of 9

Math Help - lognormal distribution

  1. #1
    Newbie
    Joined
    Apr 2009
    Posts
    6

    lognormal distribution

    Hi all, i have a few parts to a question that I'm not too sure about, and i was wondering if anyone could help me out.

    I'll start with one and add more in if i still need help.

    I'm having trouble deriving the E(X) of the log normal distribution.

    I believe i should be using moment generating functions of the normal distribution and applying a X = exp(Y) transformation? But I'm not entirely sure how to use them both to derive it.

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by anonder View Post
    Hi all, i have a few parts to a question that I'm not too sure about, and i was wondering if anyone could help me out.

    I'll start with one and add more in if i still need help.

    I'm having trouble deriving the E(X) of the log normal distribution.

    I believe i should be using moment generating functions of the normal distribution and applying a X = exp(Y) transformation? But I'm not entirely sure how to use them both to derive it.

    Thanks!
    Let X be normal with mean m and variance \sigma ^2, then e^X is lognormal.

    \mathbb{E}(e^X) = \phi_X(1), where \phi is the MGF/characteristic fn(I dont what people call them exactly)

    Since \phi_X(s) = \mathbb{E}(e^{sX}) = \text{exp}\left(\frac{s^2 \sigma ^2}2 + ms\right), we have \mathbb{E}(e^X) = \text{exp}\left(\frac{\sigma ^2}2 + m\right)
    Last edited by Isomorphism; May 23rd 2009 at 06:18 AM. Reason: thanks Moo
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,
    Quote Originally Posted by anonder View Post
    Hi all, i have a few parts to a question that I'm not too sure about, and i was wondering if anyone could help me out.

    I'll start with one and add more in if i still need help.

    I'm having trouble deriving the E(X) of the log normal distribution.

    I believe i should be using moment generating functions of the normal distribution and applying a X = exp(Y) transformation? But I'm not entirely sure how to use them both to derive it.

    Thanks!
    *without MGF* (yeah, I'm crazy about LOTUS)

    As you said, X=\exp(Y), where Y follows a normal distribution (\mu,\sigma^2). So we're looking for \mathbb{E}(\exp(Y)).

    For more convenience, we can consider Z\sim\mathcal{N}(0,\sigma^2). And then, Y follows the same distribution as Z+\mu.
    We'll calculate \mathbb{E}(\exp(Z)).
    And from the above, we'll have \mathbb{E}(\exp(Y))=e^{\mu}\cdot \mathbb{E}(\exp(Z))


    Z's pdf is \frac{1}{\sigma\sqrt{2\pi}}\cdot e^{-\frac{t^2}{2\sigma^2}}.

    Using the law of the unconscious statistician (at last I can put a name on it !!! ), we have :

    \begin{aligned}<br />
\mathbb{E}(\exp(Z)) &=\int_{\mathbb{R}} \frac{1}{\sigma\sqrt{2\pi}}\cdot e^t\cdot e^{-\frac{t^2}{2\sigma^2}} ~dt \\<br />
&=\int_{\mathbb{R}}\frac{1}{\sigma\sqrt{2\pi}} \cdot\exp\left(\frac{-t^2+2\sigma^2 t}{2\sigma^2}\right) ~dt<br />
\end{aligned}

    Complete the square : -t^2+2\sigma^2t=-(t-\sigma^2)^2+\sigma^4

    Hence :

    \begin{aligned}<br />
\mathbb{E}(\exp(Z)) &=\int_{\mathbb{R}} \left[\frac{1}{\sigma\sqrt{2\pi}}\cdot\exp\left(\frac{-(t-\sigma^2)^2}{2\sigma^2}\right)\right]\cdot e^{\sigma^2/2} ~dt \\<br />
&=e^{\sigma^2/2}\int_{\mathbb{R}} \frac{1}{\sigma\sqrt{2\pi}}\cdot\exp\left(\frac{-(t-\sigma^2)^2}{2\sigma^2}\right) ~dt<br />
\end{aligned}

    But the thing in the integral is exactly the pdf of a normal distribution (\sigma^2 ~,~ \sigma) !
    Hence this integral is 1.

    And finally, \mathbb{E}(\exp(Z))=e^{\sigma^2/2} \Rightarrow \boxed{\mathbb{E}(X)=\mathbb{E}(\exp(Y))=\exp\left  (\mu+\frac{\sigma^2}{2}\right)}


    where \phi is the MGF/characteristic fn(I dont what people call them exactly)
    As stated, this is the MGF.

    The MGF is equivalent to the Laplace transform \int e^{tx} f(x) ~dx
    The characteristic function is equivalent to the Fourier transform \int e^{itx} f(x) ~dx

    Since \phi_X(s) = \mathbb{E}(e^{sX}) = \text{exp}\left(\frac{s^2 \sigma ^2}2 - ms\right), we have \mathbb{E}(e^X) = \text{exp}\left(\frac{\sigma ^2}2 - m\right)
    The MGF of a normal is \exp\left(ms+\frac{\sigma^2s^2}{2}\right)
    For the characteristic function, just substitute t=it


    You can find the definitions of mgf and characteristic function in wikipedia, as well as the mgf, char function of any distribution.
    Last edited by Moo; May 23rd 2009 at 06:21 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Moo View Post
    As stated, this is the MGF.

    The MGF is equivalent to the Laplace transform \int e^{tx} f(x) ~dx
    The characteristic function is equivalent to the Fourier transform \int e^{itx} f(x) ~dx


    The MGF of a normal is \exp\left(ms+\frac{\sigma^2t^2}{2}\right)
    For the characteristic function, just substitute t=it
    I have lost a total of 17.5 marks for putting that negative sign last semester, and I still do it

    You can find the definitions of mgf and characteristic function in wikipedia, as well as the mgf, char function of any distribution.
    The problem is our lecture notes call it the characteristic function, last time Laurent pointed it and I saw it on Wikipedia too. I dont why they called it char function in our lecture notes. Now its confusing me a lot. Anyway I wil keep that in mind...Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Isomorphism View Post
    I have lost a total of 17.5 marks for putting that negative sign last semester, and I still do it
    Over 20 ?
    Just joking... over 100 ? That's still a lot :s

    By taking the pdf, you have very little chance of getting wrong lol! (except if, like I did in a first time, you forget the minus sign in the exp(-t^2/2sigma^2)... )
    The problem is our lecture notes call it the characteristic function, last time Laurent pointed it and I saw it on Wikipedia too. I dont why they called it char function in our lecture notes. Now its confusing me a lot. Anyway I wil keep that in mind...Thanks
    Both characterize the distribution. And you can go from one to another quite easily (though it requires justification if you want to be formal ^^)
    That's still strange ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2009
    Posts
    6
    Thanks guys! Especially Moo.

    Another question about the variance. Would you need to then use MGFs to do derive the variance? Using the 2nd moment of the normal distribution to find E(X^2) for the log normal distribution, thus having E(X^2) - m^2 for the variance.

    Though i can't get the right answer.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by anonder View Post
    Thanks guys! Especially Moo.

    Another question about the variance. Would you need to then use MGFs to do derive the variance? Using the 2nd moment of the normal distribution to find E(X^2) for the log normal distribution, thus having E(X^2) - m^2 for the variance.

    Though i can't get the right answer.
    Okay, here the MGF is more useful XD

    \mathbb{E}(X^2)=\mathbb{E}((e^Y)^2)=\mathbb{E}(e^{  2Y})=\phi_Y(2)=\exp\left(2\mu+2\sigma^2\right)
    (Y follows a normal distribution (\mu,\sigma^2))

    Hence \text{var}(X)=\mathbb{E}(X^2)-[\mathbb{E}(X)]^2=\exp\left(2\mu+2\sigma^2\right)-\exp\left(2\left(\mu+\frac{\sigma^2}{2}\right)\rig  ht)

    =\exp\left(2\mu+2\sigma^2\right)-\exp\left(2\mu+\sigma^2\right)=e^{2\mu+\sigma^2}\l  eft(e^{\sigma^2}-1\right)

    Looks good ?
    Last edited by Moo; May 23rd 2009 at 08:10 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2009
    Posts
    6
    Thanks Moo, really helpful! I found my mistake .

    Was able to do the next part too! Though i might have 2 more questions still.

    If you have alpha = Xbar as an estimator, how do you find the variance for that specific estimator? The estimator is unbiased, which was easy to show.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Apr 2009
    Posts
    6
    I figured it out with an easy estimator, but I'm now stuck on another one.

    It's a bit too long to type so i attached it I'm not sure how you use large n approximations? Actually I'm kinda pretty confused in general about it. If someone could point me in the right direction to get started, that would be awesome!

    Thanks so far, it's really helped.
    Attached Thumbnails Attached Thumbnails lognormal distribution-question.jpg  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. estimating three parameter lognormal distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 12th 2011, 12:23 AM
  2. Lognormal distribution's M.G.F?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: December 27th 2009, 02:52 AM
  3. Lognormal Distribution & Probability of Sucess
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 4th 2009, 07:21 PM
  4. Lognormal Distribution Function
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: March 25th 2009, 06:08 PM
  5. Lognormal distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: May 12th 2008, 12:32 AM

Search Tags


/mathhelpforum @mathhelpforum