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I am very new to these statistical concepts and getting confused really quickly while reading online. I am supposed to find the expectation of $\ln(X)\sim N(\mu, \sigma)$ by using the MGF.

So far I've got

$M_{\ln(X)}(t)=E[e^{t\ln(X)}]$ (By defintion) $=E[X^t]$

but I am not sure how to go from there. I saw someone online simply plugging in $1$ for $t$ and saying that's the expectation but I didn't find anything on why is that and I don't know if it's correct.

Another thing I know is that the expectation is the first derivative of the MGF evaluated at $t=0$. But I am not sure how to use that. Do I simply plug in $\ln(x)$ instead of $x$ in the MGF of $X\sim N(\mu,\sigma)$ then take the first derivative? or?

StubbornAtom
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R.Bair
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3 Answers3

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The moment-generating function of a $N(\mu, \sigma)$ random variable is $e^{\mu t + \frac{\sigma t^2}{2}}$, assuming that $\sigma$ is the variance. If you meant that $\sigma$ is the standard deviation, then $\sigma^2 $ is the variance, and then you need to put $\sigma^2$ in place of $\sigma$ above.

Given the assumed distribution of $\log X$ as you've described, it is known that $M_{\log X} (t) = e^{\mu t + \frac{\sigma t^2}{2}}$.

If you want to find the expectation of $\log X$, you can differentiate the above expression and then plug in $t = 0$.

However, if you wanted to find the expectation of $X$, then you use what you were getting at, namely that $E(X^t) = e^{\mu t + \frac{\sigma t^2}{2}}$, and then you can certainly plug in $t = 1$ to find the answer.

Novice
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If $\ln X\sim\operatorname N(\mu,\sigma^2)$ then $\operatorname E(\ln X)$ is just $\mu.$

But I have to suspect you meant the expected value of $X$ if $\ln X \sim\operatorname N(\mu,\sigma^2).$ You should work on expressing things like that more clearly.

\begin{align} \operatorname E(X) & = \int_0^\infty x f_X(x)\, dx \\[8pt] & = \int_{-\infty}^{+\infty} e^u f_U(u)\, du \text{ where } u = \ln x, \text{ so that } x = e^u, \\[8pt] & = \int_{-\infty}^{+\infty} e^u \cdot \frac 1 {\sqrt{2\pi}} e^{-(1/2)((u-\mu)/\sigma)^2} \, \left( \frac {du} \sigma \right) \\[8pt] & = \int_{-\infty}^{+\infty} e^{\mu + \sigma z} \cdot \frac 1 {\sqrt{2\pi}} e^{-z^2/2} \, dz \\[8pt] & = e^\mu \cdot \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} \exp\left( \frac{-1}2 (z^2 - 2\sigma z) \right) \, dz \\[8pt] & = e^\mu \cdot \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} \exp\left( \frac{-1}2 (z^2 - 2\sigma z + \sigma^2) \right) \, dz \cdot \exp\left( \frac{\sigma^2} 2 \right) \\ & \qquad \text{The reason we can pull out } \exp(\sigma^2/2) \text{ is that} \\ & \qquad \text{it does not change as $z$ goes from } {-\infty} \text{ to } {+\infty}. \\[8pt] & = e^\mu \cdot \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{(-1/2)(z-\sigma)^2} \, dz \cdot e^{\sigma^2/2} \\[8pt] & = e^{\mu+\sigma^2/2} \cdot 1. \end{align}

Michael Hardy
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  • Thanks for the detailed answer. Yes, that's what I meant, with a slight difference. $lnX\thicksim N(\mu,\sigma)$, and not $\sigma^{2}$. I am wondering why you and the other people answering the question changed it to $\sigma^{2}$? – R.Bair Aug 11 '20 at 05:17
  • @ChristopherToni : $\sigma^2$ is the variance. $\qquad$ – Michael Hardy Aug 11 '20 at 06:32
  • @R.Bair : Do you have in mind that $\sigma$ is the standard deviation or that $\sigma$ is the variance? $\qquad$ – Michael Hardy Aug 11 '20 at 06:33
  • I am not sure to be honest. It’s a homework question and it simile says “let $ln(X)\thicksim N(\mu,\sigma)$. Find EX.” How would I know in this case? – R.Bair Aug 11 '20 at 14:29
  • @R.Bair Couldn't you just write something like "if the variance is sigma, then the answer is this, but if the variance is supposed to be sigma squared, then the answer is that"? – Novice Aug 11 '20 at 15:52
  • Probably what was intended was that $\sigma$ is the standard deviation and $\sigma^2$ is the variance, and somewhat less-than-standard notation was used. – Michael Hardy Aug 11 '20 at 18:36
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For $\log(X)\sim\mathcal{N}(\mu,\sigma^2)$, you find $$E[\log(X)]=\mu$$ But as you have mentioned, it holds

$$E[X]=\mathcal{M}_{\mathcal{N}(\mu,\sigma^2)}(1)=\exp\Big(\mu+\frac{\sigma^2}{2}\Big)$$

which is of course the expectation of $X$. (Observe that the moment generating function is only a function of $t$, i.e. if you put any $t$ in it, you get a real value. See https://www.youtube.com/watch?v=L8NS33q3YYc)

user408858
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