Suppose the conditional variable is distributed as $X\mid F_t\sim\mathcal{N}(0, \alpha^2)$ and $X\sim\mathcal{N}(0,\beta^2)$.
I want to determine $E[\exp(X)\mid F_t]$. I know that the unconditional expectation is given by $E[\exp(X)]=\exp(\frac{\beta^2}{2})$, now my question is; for the conditional expectation can I simply conclude that $E[\exp(X)\mid F_t]=\exp(\frac{\alpha^2}{2})$?
Intuitively the conditional expectation just provides additional information regarding the generating sigma algebra, so my gut instinct says that the conditional expectation is correct. Although I would not be writing this question if if I was convinced, hence could someone provide some help/guidance on this problem.
Thanks for reading.
Best Answer
\begin{align*} \mathbb E\left[e^X \mid F_t\right] &= \int\limits_{x=-\infty}^\infty e^x f(X=x|F_t)dx \end{align*} Since $X\mid F_t\sim\mathcal{N}(0, \alpha^2)$, $f(X\mid F_t) = \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left\{-\frac{x^2}{2\alpha^2}\right\}$
Substituting this back, \begin{align*} \mathbb E\left[e^X \mid F_t\right] &= \int\limits_{x=-\infty}^\infty exp(x) \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left(-\frac{x^2}{2\alpha^2}\right)dx\\ &= \int\limits_{x=-\infty}^\infty \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left(x-\frac{x^2}{2\alpha^2}\right)dx\\ &= \int\limits_{x=-\infty}^\infty \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left(\frac{\alpha^4-\alpha^4+2\alpha^2x-x^2}{2\alpha^2}\right)dx\\ &= \int\limits_{x=-\infty}^\infty \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left(\frac{\alpha^2}{2}-\frac{(x-\alpha^2)^2}{2\alpha^2}\right)dx\\ &= \exp\left(\frac{\alpha^2}{2}\right)\int\limits_{x=-\infty}^\infty \frac{1}{ \alpha\sqrt{2 \pi}} \exp\left(-\frac{(x-\alpha^2)^2}{2\alpha^2}\right)dx\\ \end{align*} The term on the right is the pdf of a normal variable with standard deviation $\alpha$ and mean $\alpha^2;\,\sim\mathcal N(\alpha^2,\alpha)$. The integral of any pdf over it's domain (in this case the reals) is 1.
Thus the above expression is \begin{align} \mathbb E\left[e^X \mid F_t\right] &= \exp\left(\frac{\alpha^2}{2}\right)\times 1\\ &= \exp\left(\frac{\alpha^2}{2}\right) \end{align} which is what you expected