I am trying to proof the well known formula of $E[\exp\left(X\right)] = \exp\left(\mu + \frac{1}{2}\sigma^2\right)$ when $ X \sim N(\mu, \sigma^2)$ by using Taylor expansion ($E[\cdot]$ denotes expectation here). I know it can be proved via integration of the product of the normal density and the exponential functions, but i think it should be also possible via Taylor expansion given the fact that all moments of the normal distribution up to the second one are zero.
This is the reasoning i apply. By Taylor we have that
$E[f(X)] = E\left[f(x_0) + f^{'}(x_0)(X-x_0) + \frac{1}{2} f^{''}(x_0)\left(X-x_0\right)^2 + \frac{1}{6} f^{'''}(x_0) \left(X-x_0\right)^3 + \cdots \right]$
applying this equation to $f(x)=\exp(x)$ and $x_0=\mu$ we have
\begin{eqnarray} E[\exp(X)] = E\left[\exp(\mu) + \exp(\mu)(X-\mu) + \frac{1}{2} \exp(\mu)\left(X-\mu\right)^2 + \frac{1}{6} \exp(\mu) \left(X-\mu\right)^3 + \cdots \right] \\ = \exp(\mu) + \exp(\mu)E[(X-\mu)] + \frac{1}{2} \exp(\mu)E\left[\left(X-\mu\right)^2\right] + \sum_{i=3}^{\infty} \frac{1}{i!} \exp(\mu) E\left[\left(X-\mu\right)^i\right] \end{eqnarray}
and taking into account that $\mu = E[X]$ and $E[(X-\mu)^i] =0$ when $i > 2$ we have that
$E[\exp(X)] = \exp(\mu) + \frac{1}{2} \exp(\mu)\sigma^2 $
... Why i do not get $E[\exp\left(X\right)] = \exp\left(\mu + \frac{1}{2}\sigma^2\right)$. What is wrong in the reasoning?
Thanks in advance.
Thanks for the comment, that was a terrible mistake.
Now I could finish the proof:
Given the fact that $E[\left(X-\mu \right)^j] = 0$ if $j$ is odd, and $E[\left(X-\mu \right)^j] = (j-1)!! \sigma^{j}$ if $j$ is even ($j!!$ is the double factorial of $j$), we have that:
\begin{eqnarray} E[\exp(X)] &=& \exp(\mu)\left( 1 + \sum_{j=1}^{\infty} \frac{1}{(2j)!} \left(2j-1 \right)!! \sigma^{2j} \right) \\ &=& \exp(\mu)\left( 1 + \sum_{j=1}^{\infty} \frac{1}{(2j)!} \frac{\left(2j-1 \right)!}{2^{j-1} (j-1)!} \sigma^{2j} \right) \\ &=& \exp(\mu)\left( 1 + \sum_{j=1}^{\infty} \frac{1}{(2j)} \frac{1}{2^{j-1} (j-1)!} \sigma^{2j} \right) \\ &=& \exp(\mu)\left( 1 + \sum_{j=1}^{\infty} \frac{1}{(j! 2^j)}\sigma^{2j} \right) \\ &=& \exp(\mu)\left( 1 + \sum_{j=1}^{\infty} \frac{1}{j!}\left(\frac{\sigma^{2}}{2} \right)^j \right) \\ &=& \exp(\mu) \left( \exp\left(\frac{1}{2}\sigma^2 \right) \right) \\ &=& \exp \left( \mu + \frac{1}{2}\sigma^2 \right) \end{eqnarray}