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I'm having trouble deriving an expression for the expected value for the lognormal distribution. I've tried the standard approach of computing $\int_{\mathbb{R^+}}xf_X(x)\,\mathrm{d}x$ for non-negative variables:

$$\int_0^{\infty} \frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\frac{\ln(y)-\mu}{\sigma}\right)^2\right)\,\mathrm{d}y$$

which is beyond me.

I've tried looking into moment generating functions, of which my knowledge is lacking, but stumbled upon a question claiming (and proving) that there is no such function. (link)

I've looked at a similar question (same, really) (link), but I'm afraid I don't undestand the accepted answer. It seems to relate the moment generating function of the normal distribution to the lognormal one, which didn't exist?

So how does one extract the expected value for the lognormal distribution, given the moment generating function of another(/the normal) distribution?

Bonus question: Is this last method the most natural approach (yes/no), or is it possible to find the expected value using the first approach with some clever trick (yes/no).

Frank Vel
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  • This is technically a duplicate question, but since I don't understand the answer to the question, this seeks to get an explanation to that answer or a more thorough explanation. – Frank Vel Aug 29 '17 at 09:38
  • I have found a solution to the first approach here. – Frank Vel Aug 29 '17 at 09:54
  • Is this answer helfpful for you? – drhab Aug 29 '17 at 10:01
  • Btw, your integral must go over $(0,\infty)$ and not over $(-\infty,\infty)$. – drhab Aug 29 '17 at 10:14
  • The bottom line is, make use of the relationship between normal and log-normal. If $X$ is normally distributed, then $Y = e^X$ is log-normally distributed. From here if you are familiar with the calculations with normal distribution, then many related quantities of log-normal can be computed in this way. – BGM Aug 29 '17 at 10:15
  • @drhab I see I have some reading to do on MGF. I'll fix the limits, I didn't know about the alternate definition for non-negative variables. – Frank Vel Aug 29 '17 at 10:34

2 Answers2

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Standard method to find expectation(s) of lognormal random variable.

1)

Determine the MGF of $U$ where $U$ has standard normal distribution.

This comes to finding the integral:$$M_U(t)=\mathbb Ee^{tU}=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu}e^{-\frac12u^2}du=e^{\frac12t^2}$$

2)

If $Y$ has lognormal distribution with parameters $\mu$ and $\sigma$ then it has the same distribution as $e^{\mu+\sigma U}$ so that: $$\mathbb EY^{\alpha}=\mathbb Ee^{{\alpha}\mu+{\alpha}\sigma U}=e^{{\alpha}\mu}\mathbb Ee^{{\alpha}\sigma U}=e^{{\alpha}\mu}M_U({\alpha}\sigma)=e^{{\alpha}\mu+\frac12{\alpha}^2\sigma^2}$$

drhab
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  • Could you please elaborate the last equality of method 1 because I thought $\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu}e^{-\frac12u^2}du=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu-\frac12u^2}=\frac1{\sqrt{2\pi}}e^{tu-\frac12u^2}$? – Nemo Jan 26 '20 at 11:09
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    @Nemo $$M_{U}\left(t\right)=\mathbb{E}e^{tU}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tu}e^{-\frac{1}{2}u^{2}}du=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left(u-t\right)^{2}+\frac{1}{2}t^{2}}du=$$$$e^{\frac{1}{2}t^{2}}\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left(u-t\right)^{2}}du=e^{\frac{1}{2}t^{2}}$$ – drhab Jan 26 '20 at 13:23
  • Thanks, @drhab for showing me the rearrangement of the exponent. However, how could I tell that $\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac12(u-t)^2}du$ is evaluated to 1, although it is somehow similar to the pdf of normal distribution (it is lack of $\mu$ and $\sigma$), or is it because it is the pdf of a random variable with mean of t and variance of 1? – Nemo Jan 27 '20 at 00:03
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    @Nemo Yes. It is exactly the PDF of a random variable with normal distribution having mean $t $ and variance $1$. So the integral over it equals $1$. – drhab Jan 27 '20 at 07:28
  • Thanks for your confirmation, @drhab. I simply can't comprehend this issue: you have concretely shown that lognormal random variable (same as distribution?) has mgf but the link https://math.stackexchange.com/questions/503165/proving-that-the-lognormal-distribution-has-no-moment-generating-function provided in the OP says mgf of lognormal does not exist. The information from that link was wrong, wasn't it? – Nemo Jan 28 '20 at 00:00
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    @Nemo I have not shown that the lognormal random variable has mgf. What I did was finding the mgf of standard normal distribution and on base of that result I showed how you can calculate several expectations of the lognormal random variable on a neat way. That is definitely not the same as showing that the lognormal random variable has a mgf. As shown in the linked question the lognormal random variable does not have a mgf, but that does not imply that expectations of $Y^{\alpha}$ do not exist. – drhab Jan 28 '20 at 09:16
  • Thanks @drhab for your clarification. Much appreciated. – Nemo Jan 28 '20 at 10:31
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Hint:

By the substitution $y=e^z$, you transform to

$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} e^z\exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2\right)e^z\,\mathrm{d}z=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\infty} \exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2+2z\right)\,\mathrm{d}z$$ which you can reduce to a standard Gaussian integral by shifing the variable, giving the value $1$.