Expected value of $X$ if $\ln(X)\sim N(\mu, \sigma)$ 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?
 A: 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.
A: 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}
A: 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)
