Expectation of $X_T^4$ when $X_T$ is log-normally distributed

Question

Let $X_T$ be a random variable with $$\ln(X_T) \sim \mathcal{N}\left(\ln(x)+\frac{T-t}{2},T-t\right).$$ What is $\mathbb{E}\left(X_T^4\right)$?

Answer 1

Hint

I will let $X_T=X$ and write $\ln X\sim \mathcal N(\mu,\sigma^2)$ in place of your expressions, for simplicity.

The pdf of the lognormal distribution is $(x\sigma\sqrt{2\pi})^{-1}\exp(\frac{-(\ln x-\mu)^2}{2\sigma^2})$ for $x>0$. Therefore, $$ E[X^4]=\int_0^\infty x^4\cdot \frac1{x\sigma\sqrt{2\pi}}\cdot \exp\left(\frac{-(\ln x-\mu)^2}{2\sigma^2}\right)\,dx $$ To compute this, start with the change of variables $z=\ln x$, so that $dz=dx/x$. Then $$ E[X^4]=\frac1{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{4z}\cdot \exp\left(\frac{-(z-\mu)^2}{2\sigma^2}\right)\,dx=\frac1{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(4z-\frac{(z-\mu)^2}{2\sigma^2}\right)\,dx $$ There is a quadratic function of $z$ inside the $\exp()$. The next step is to complete the square so that function becomes $\exp(\frac{-(z-c)}{2\sigma^2})$ for some $c$. Once you have done this, the integrand just becomes a shifted version of the normal cdf, which integrates to one. The result is just the constant you had to multiply by in order to complete the square.

Answer 2

Given:
You are given that $$ \ln(X_T) \sim \mathcal{N}\left(\ln(x) + \dfrac{T-t}{2}, T-t \right) \;. $$

Simplifications:
To begin with, let us define $$ Y := \ln(X_T) \;, \qquad \mu := \ln(x) + \dfrac{T-t}{2} \;, \qquad\text{and}\qquad \sigma := T - t \;. $$ Then the given expression is equivalent to the much simpler expression $Y \sim \mathcal{N}(\mu, \sigma)$. For this, we already know that $$ \mathbb{E}(Y) = \color{green}{\mu} \;, \qquad \mathbb{SD}(Y) = \color{brown}{\sigma} \;, \qquad\text{and}\qquad \mathbb{E}(X_T) = \mathrm{e}^{\color{green}{\mu} + \frac{1}{2}\color{brown}{\sigma}^2} \;. $$ To get from $\mathbb{E}(X_T)$ to $\mathbb{E}(X_T^4)$, we need an additional step.

Identities:
Notice that $\ln(X_T^c) = cY$ for every $c > 0$. Using the identities $$ \mathbb{E}(cY) = c\mathbb{E}(Y) \qquad\text{and}\qquad \mathbb{SD}(cY) = c\mathbb{SD}(Y) \;, $$ we obtain $\ln(X_T^c) = cY \sim \mathcal{N}(c\mu, c\sigma)$. From this it follows that $$ \mathbb{E}(cY) = \color{green}{c\mu} \;, \qquad \mathbb{SD}(cY) = \color{brown}{c\sigma} \;, \qquad\text{and}\qquad \mathbb{E}(X_T^c) = \mathrm{e}^{\color{green}{c\mu} + \frac{1}{2}(\color{brown}{c\sigma})^2} \;. $$

Final result:
Now we just need to set $c = 4$, to get the final result

$$ \mathbb{E}(X_T^4) = \mathrm{e}^{4\mu + \frac{1}{2}(4\sigma)^2} = \underline{\underline{x^4 \cdot \mathrm{e}^{2(T-t) + 8(T-t)^2}}} \;. $$