5
$\begingroup$

If X is a random variable with a normal distribution, then Y = exp(X) has a log-normal distribution.

Likewise if X is a random variable with a binomial distribution, then Y = exp(X) has a log-binomial distribution.

My question
In case of the log-normal distribution mean and variance are well known, for the log-binomial distribution I can't find any references.

$\endgroup$
1
  • 3
    $\begingroup$ Hint: $\mathbb E Y = m_X(1)$ where $m_X(t)$ is the moment-generating function of $X$. Now, what about the variance? $\endgroup$
    – cardinal
    Apr 15 '12 at 19:20
10
$\begingroup$

We can use an entirely analogous technique to the one typically used to calculate the moments of a lognormal.

In particular, note that if $\newcommand{\E}{\mathbb E}X \sim \mathrm{Bin}(n,p)$ and $Y = e^X$, then $Y^k = e^{k X}$. But, $\E e^{kX} = m_X(k)$ where $m_X(t)$ is the moment-generating function of $X$ evaluated at $t$.

Hence, $$ \E e^{k X} = m_X(k) = (1 - p + p e^k)^n \>, $$ and so $$ \E Y = m_X(1) = (1 + (e-1)p)^n $$ and $$ \mathrm{Var}(Y) = m_X(2) - (m_X(1))^2 = (1+(e^2-1)p)^n - (1+(e-1)p)^{2n} \>. $$

Other moments of $Y$ can be computed in a similar fashion.

$\endgroup$
1
$\begingroup$

See: Parameter Estimation and Goodness-of-Fit in Log Binomial Regression, L. Blizzard and D. W. Hosmer, Biometrical Journal Volume 48, Issue 1, pages 5–22, February 2006

$\endgroup$
1
  • 3
    $\begingroup$ You should explain why this article helps solving the problem. Please provide some details. $\endgroup$ Feb 22 '13 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.