I am reading a paper and got stuck on this simple equation:
$$\mathbb{E}_t[e^{-cS_T}]$$ where $dS_t=\sigma W_t$ with $W_t$ standard 1 dimensional Brownian motion, $S_t=s$ and c some constant.
I tried this as follows:
$$\mathbb{E}_t[e^{-cS_T}]=\mathbb{E}_t[e^{-c(\sigma(W_T-W_t)+s)}]=e^{\frac{1}{2}c^2\sigma^2(T-t)^2}e^{cs}$$ Because $$S_T=\sigma(W_T-W_t)+s \sim N(s,\sigma(T-t))$$ However, the paper states the solution as $$e^{\frac{1}{2}c^2\sigma^2(T-t)}e^{cs}$$ Why is there not a square in their solution? What am I doing wrong?
I was told that there is another possibility to compute the expectation: Let $Z_t=e^{(-\alpha t-\beta W_t)}$ and find $\alpha , \beta$ such that $Z_t$ is a martingale. But I don't understand how I would use that to compute the expectation. Would appreciate if someone could give me a hint there as well. Thanks!
Let $X$ be a Gaussian random variable with mean $m$ and variance $\varrho^2$, then
$$\mathbb{E} \exp(\lambda X) = \exp \left( \lambda m + \frac{1}{2} \lambda^2 \varrho^2 \right). \tag{1}$$
holds for any $\lambda \in \mathbb{C}$. From
$$S_T = \sigma \underbrace{(W_T-W_t)}_{\sim N(0,T-t)} + s \sim N(s, \sigma^2 (T-t))$$
it therefore follows that
$$\mathbb{E}\exp(-c S_T) = \exp \left( -c s + \frac{1}{2} (-c)^2 \sigma^2 (T-t) \right) = \exp \left(-cs + \frac{1}{2} c^2 \sigma^2 (T-t) \right).$$