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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.

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    $\begingroup$ 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
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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.

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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

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We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.

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

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