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I need to calculate the first and second moment of a log normal with mean=$\log S_0-(\mu-\frac{\sigma^2}{2})(t-t_0)$ and variance=$\sigma^2(t-t_0)$. I know that the answer should be $E[X]=S_0e^{\mu(t-t_0)}$. I fill in the equation. where I put $(t-t_0)=\Delta t$: $$\int_{0}^{\infty}Sf(S;t-t_0,S_0)dS=\int_{0}^{\infty}S\frac{1}{S\sigma \sqrt{2\pi\Delta t}}e^{-\frac{(\log S-\log S_0-(\mu-\frac{\sigma^2}{2})(\Delta t))^2}{2\sigma^2 \Delta t}}dS$$

Then I take $y=\log\frac{S}{S_0}$ $$\int_{0}^{\infty}\frac{1}{\sigma \sqrt{2\pi\Delta t}}e^{-\frac{(y-(\mu-\frac{\sigma^2}{2})(\Delta t))^2}{2\sigma^2 \Delta t}}dy$$

I think I need to make a quadratic part above the $e$ so I can make that integral 1 as it is a density function.

Can someone help me on how to do this?

Eran
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  • Can you not use the formulas for the two moments of the log-normal distribution? If $X\sim\operatorname{LN}(\mu_{x},\sigma_{x}^2)$ then $E[X^n] = e^{n\mu_x + n^2\sigma_x^2/2}$. Just fill in with $\mu_{x} = \log S_0 - (\mu-\sigma^2/2)(t-t_0)$ and $\sigma_{x}^2 = \sigma^2(t-t_0)$. – Therkel Mar 29 '17 at 12:41
  • Yes but I need to derive it, but got it for the first moment know, with http://math.stackexchange.com/questions/628681/how-to-compute-moments-of-log-normal-distribution – Eran Mar 29 '17 at 16:25
  • Can you please clarify? The accepted answer in the linked question shows you exactly how to derive the result in my previous comment. Does it not? – Therkel Mar 29 '17 at 16:37
  • Yes I know, but the it is a bit different as I needed to change the variables here, and the part that was hard there as well: the manipulation of the exponential in a specific way. That was what I was looking for. I already have seen the end result which was you answer . – Eran Mar 30 '17 at 07:15

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