Expected Value of Exponential Function of Stochastic Process: $E[e^{X_t}]$ I am still getting used to expectations and stochastic processes. Supposed we have a stochastic process defined by
$$dX_t = \mu(t,X_t)dt+\sigma(t,X_t)dW_t$$
where $W_t$ is a Wiener process. The simple thing I need to do is evaluate the following expression for some time $T>t$.
$$E_{t,x}[e^{X_T}]$$
Would I be correct in assuming the following logic?
$$E_{t,x}[e^{X_T}]=e^{E_{t,x}[X_T]}$$
Which then we can use Itô's Lemma to find $E_{t,x}[X_T]$. First we define $f(t,x)=x$. Now we have the stochastic differential. Here we use the facts that $\frac{\partial f}{\partial t}=0$ and $\frac{\partial^2 f}{\partial x^2}=0$
$$df(t,X_t)=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$$
Now we integrate
$$\int_t^T df(t,X_t)=\int_t^T\mu(t,X_t)dt+\int_t^T\sigma(t,X_t)dW_t$$
Which implies this (the expected value of the $dW_t$ integral is zero).
$$f(T,X_T)=f(t,X_t)+\int_t^T\mu(t,X_t)dt$$
Finally we take expectation.
$$E_{t,x}[f(T,X_T)]=E_{t,x}[X_T]=f(t,X_t)+ \int_t^T E_{t,x}[\mu(t,X_t)]dt$$
So then this leaves me with the final representation of the expectation that I am looking for.
$$E_{t,x}[e^{X_T}]=e^{E_{t,x}[X_T]}=e^{f(t,X_t)+ \int_t^T E_{t,x}[\mu(t,X_t)]dt}$$
I guess then we just hope that the $\mu$ function is easy enough to analyze. This just still doesn't feel so comfortable to me, I was wondering if I'm doing this right or if I'm missing something here.
Thanks.
 A: As pointed out in the comments above, it is not correct to assume that $\mathbb{E}[e^X]=e^{\mathbb{E}[X]}$.
The easiest counterexample is Normal vs. Lognormal distribution. If $X\sim N(\mu,\sigma)$, then $Y:=e^X$ is Lognormal by definition (i.e. see definition of Lognormal distribution).
The expected value of $Y$ is $\mathbb{E}[e^{X}]=e^{\mu+0.5\sigma^2}\neq e^{\mu}=e^{\mathbb{E}[X]}$.
You can either take the above as a "fact" that arises from the definition of Lognormally distributed random variable (i.e. you can memorize what the mean of a Lognormally distributed random variable looks like), or we can derive it by deriving the distribution of $Y$ from scratch:
$$F_Y(y)=\mathbb{P}(Y\leq y)=\mathbb{P}(e^{X}\leq y)=\mathbb{P}(X\leq ln(y))=\int_{-\infty}^{ln(y)}f_X(h)dh$$
Now:
$$f_Y(y):=\frac{\partial}{\partial y}\left(F_Y(y)\right)=\frac{\partial}{\partial y}\left(\mathbb{P}(X\leq ln(y))\right)=\frac{\partial}{\partial ln(y)}\left(\int_{-\infty}^{ln(y)}f_X(h)dh\right)\frac{\partial ln(y)}{\partial y}=\\=f_X(ln(y))\frac{1}{y}=\\=\frac{1}{\sqrt{2\pi}}e^{\frac{-(ln(y)-\mu)^2}{2\sigma^2}}\frac{1}{y}$$
Taking the expectation:
$$\mathbb{E}[Y]:=\int_{-\infty}^{\infty}\left(y f_Y(y)\right)dy=\\=\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(ln(y)-\mu)^2}{2\sigma^2}}\right)dy$$
Taking the substitution $y=e^h$, we get:
$$\mathbb{E}[Y]=\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(h-\mu)^2}{2\sigma^2}}\right)e^hdh$$
Now focusing on the exponents:
$$exp\left\{\frac{-(h-\mu)^2}{2\sigma^2}\right\}exp\left\{h\right\}=exp\left\{\frac{-h^2+2\mu h -\mu^2}{2\sigma^2}\right\}exp\left\{\frac{2\sigma^2h}{2 \sigma^2}\right\}=\\=exp\left\{\frac{-h^2+2h(\mu+\sigma^2) - (\mu+\sigma^2)^2 + (\mu+\sigma^2)^2 -\mu^2}{2\sigma^2}\right\}=\\=exp\left\{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}\right\}exp\left\{\frac{(\mu+\sigma^2)^2 -\mu^2}{2\sigma^2}\right\}=\\=exp\left\{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}\right\}exp\left\{\mu + 0.5\sigma^2\right\}$$
Plugging this back into the integral, we get:
$$\mathbb{E}[Y]=exp\left\{\mu + 0.5\sigma^2\right\}\int_{-\infty}^{\infty}\left(\frac{1}{\sqrt{2\pi}}e^{\frac{-(h-(\mu+\sigma^2))^2}{2\sigma^2}}\right)dh=\\=exp\left\{\mu + 0.5\sigma^2\right\}$$
Because the expression in the integral is the PDF of a Normally distributed random variable (with mean $\mu+\sigma^2$): which integrates to 1.
Now to your problem:
$$X_t=X_0+\int_{h=0}^{h=t}\mu(X_h,h)dh+\int_{h=0}^{h=t}\sigma(X_h,h)dW_h$$
If $\mu(X_h,h)=\mu X_h$ with $\mu$ being a constant and $\sigma(X_h,h)=\sigma X_h$ with $\sigma$ being a constant, then $X_t$ is Geometric Brownian Motion, which has the well known solution:
$$X_t=X_0exp\left\{\mu t -0.5 \sigma^2 t + \sigma W_t\right\}$$
From the above, we can see directly that $X_t$ is log-normally distributed. So that:
$\mathbb{E}[X_t]=X_0e^{\mu t}$
If you want to take the conditional expectation $\mathbb{E}_t[X_T]=\mathbb{E}[X_T|\mathcal{F}_t]$, then:
$$\mathbb{E}[X_T|\mathcal{F}_t]=\mathbb{E}\left[X_t exp\left\{\mu (T-t) -0.5 \sigma^2 (T-t) + \sigma W(T-t)\right\}\right]=\\=X_te^{\mu (T-t)}$$
Edit: if we cannot assume that the process is GBM (and therefore know its distribution), but we have to stick with the general $\mu(X_t,t)$ and $\sigma(X_t,t)$ functions (that, we must assume, are at least square-integrable, for $X_t$ to be an Ito Process), we can apply Ito's lemma as follows:
Ito's Lemma states that any twice differentiable function $F()$ of $X_t$ and $t$ follows the process:
$$F(X_t,t)= F(X_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}\mu(X_h,h)+\frac{1}{2}\frac{\partial^2 F}{\partial X^2}\sigma(X_h,h)^2\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}\sigma(X_h,h)\right)dW_h $$
We can apply the above formula to $e^{X_t}$ to get:
$$e^{X_t}=e^{X_0}+\int_{h=0}^{h=t}\left(0+e^{X_h}\mu(X_h,h)+\frac{1}{2}e^{X_h}\sigma(X_h,h)^2\right)dh + \int_{h=0}^{h=t}\left(e^{X_h}\sigma(X_h,h)\right)dW_h$$
Now we could attempt to take an expectation of the above: you are correct in your question to say that the expectation will "kill" the Ito Integral (because of the martingale property of the Ito integral, its expectation is equal to zero), but unless we know what the functions $\sigma(X_h,h)$ and $\mu(X_h,h)$ actually are, we won't be able to get any proper answer anyway, because we won't be able to simplify the expectation over the Riemann integral.
