Show $\mathbb E[e^{4W_t-8t}(W_t-4t)^4]=3t^2$ without Girsanov

Question

I would like to know if it is possible to prove the identity $\mathbb E[e^{4W_t-8t}(W_t-4t)^4]=3t^2$ without the use of a measure change and Girsanov Theorem? I only know the solution by using the exponential martingale for the first term and then simplify terms under the new measure.

Answer 1

Since $W_t$ is a normal random variable with mean $0$ and variance $t$, we have $$ \mathbb{E}[e^{4W_t-8t}(W_t-4t)^4]=\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}(x-4t)^4e^{4x-8t-\frac{x^2}{2t}}\;dx$$ $$ =\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}(x-4t)^4e^{-\frac{1}{2t}(x-4t)^2}\;dx=\frac{1}{\sqrt{2\pi t}}\int_{-\infty}^{\infty}u^4e^{-\frac{u^2}{2t}}\;du$$ $$=\frac{t^2}{\sqrt{2\pi}}\int_{-\infty}^{\infty}y^4e^{-\frac{y^2}{2}}\;dy=3t^2$$

using the fact that the fourth moment of a standard normal random variable is $3$.

Answer 2

For arbitrary real $a$, $$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}e^{4x-2a\sigma^2}(x-a\sigma^2)^4 \, dx = e^{(8-2a)\sigma^2}\sigma^4 \left( (a-4)^4\sigma^4 + 6(a-4)^2 \sigma^2 + 3\right) $$ The relation you want to prove is the special case of $a=4$.