We have that $\{W_t:t\geq 0\}$ is a Brownian motion, and $\{N_t:t\geq 0\}$ is a rate one poisson process which is independent of the Brownian motion. We must show that $$\mathbb{E}[W_{N_t}^2]=t$$ I figured there are two ways to do this, but they yield different answers. Why is the one of them (or both) incorrect?
Method 1: Using the tower property we have that $$\mathbb{E}[W_{N_t}^2]=\mathbb{E}[\mathbb{E}[W_{N_t}^2|\mathscr{F}_{N_t}]]=\mathbb{E}[N_t]=t$$ Where $\mathscr{F}_{N_t}=\sigma\{W_s,0\leq s\leq N_t\}$
Method 2: We can use the fact that $$W_{N_t}\sim N(0,N_t)$$ Which would mean that we have $$\mathbb{E}[W_{N_t}^2]=Var(W_{N_t})=N_t$$
Obviously there is something going wrong in method 2, but what is it? Didn't we use pretty much the same trick in method 1, except there we conditioned on the filtration? Why does this ''matter'' here? Does $W_{N_t}$ not have this distribution?
Any help is appreciated.