$\mathbb{E}[e^{2B_t -t}] = e^{2t-t}$ - why?

Question

Let $B_t$ be brownian motion at time t.

I know we can use properties of MGF, but I get a different answer:

in general, for $X \sim N(\mu, \sigma^2)$ $$\mathbb{E}[\exp(\alpha X)] = \exp(\mu \alpha + \cfrac{\alpha^2 \sigma^2}{2})$$

So in our case, $X = B_t - t$

$$X \sim N(-t, 0)$$

since $\mathbb{E}[X] = 0 -t$ and $Var(B_t - t) = t-t=0 \space (\text{possible fail here?})$

Answer 1

Since $B_t$ is a Brownian motion, we have $\mathbb{E}[B_t]=0$ and $Var[B_t]=t$. Now, let's observe $\mathbb{E}[e^{2B_t-t}]$.

First, we can write

$$\mathbb{E}[e^{2B_t-t}]=e^{-t}\mathbb{E}[e^{2B_t}]$$

Next, recall that for $X\sim N(\mu,\sigma^2)$, we have $\mathbb{E}[e^{\alpha X_t}]=e^{\mu \alpha+\frac{\alpha^2\sigma^2}{2}}$.

Thus, for $X_t=B_t\sim N(0,t)$, we have

$$\mathbb{E}[e^{2 B_t}]=e^{0\times 2+\frac{2^2 t}{2}}=e^{2t}$$

Putting it all together, we have

$$\mathbb{E}[e^{2B_t-t}]=e^t$$

as expected!

Answer 2

The variance of $B_t$ is $t$. Shifting its mean by $t$ does not effect its variance. So $Var(B_t - t) = t$.