Let $W = (W_t)$, $t≥0$ be a Brownian motion. Let $s, t ≥ 0$. Then:
i) Show that $E(W_sW_t)=min(s,t)$
ii) Show that $E|W_t−W_s|^{4} =3|t−s|^{2}$
iii) Find the distribution function of the random variable $X = W_t + W_s$.
I've been able to show $E(W_sW_t)=min(s,t)$ given that $s<t$, however I'm struggling to understand how to do parts ii and iii. Any help would be great, thank you
Without loss of generality assume $s<t$. For the second part, by definition of Brownian motion $W_t - W_s \sim N(0, t-s)$. Now the result is simply the fourth moment of a normal distribution, since a centred Gaussian $X$ has $\mathbb{E}[X^4] =3 \sigma^2$. This is a direct calculation using the probability mass function and integrating by parts (just find $\int x^4 e^{-\frac{x^{2}}{2}}dx$).
For the last part, $W_t + W_s = 2W_s + (W_t - W_s)$. This is a sum of two independent Gaussians, hence is normally distributed with mean $0$ and variance $4s + t-s=3s+t $.