0
$\begingroup$

I'm trying to approximate CDF of the integral $$\frac{1}{T}\int_0^T e^{\sigma W_t+\left(r-\frac{\sigma^2}{2}\right)t}dt,$$

where $W_t$ is the Wiener process, i.e. $W_t\sim N(0,t)$.

For this I use right Riemann sum: $$\frac{1}{T}\int_0^T e^{\sigma W_t+\left(r-\frac{\sigma^2}{2}\right)t}dt\approx\frac{1}{N}\sum_{k=1}^N e^{\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}}$$

It is known that $$\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}\sim N\left(\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}, \sigma^2\frac{Tk}{N}\right)$$

The $m$-th moment of the random variable $e^{\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}}$ is $e^{m\frac{Tk}{N}\left(r-\frac{\sigma^2}{2}\right)+\frac{m^2\sigma^2}{2}\frac{Tk}{N}}$.

The first moment is $e^{\frac{Tk}{N}r}$.

The second moment is $e^{\frac{Tk}{N}\left(2r+\sigma^2\right)}$.

The third moment is $e^{\frac{Tk}{N}\left(3r+3\sigma^2\right)}$.

Now, I'm trying to apply the Berry-Esseen theorem: https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands (we need to subtract expected values to make expected value equal zero):

$$\left\lvert\frac{\sum_{k=1}^N e^{\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}}-\sum_{k=1}^N e^{\frac{Tk}{N}r}}{\sqrt{\sum_{k=1}^N e^{\frac{Tk}{N}\left(2r+\sigma^2\right)}}}-\Phi(x)\right\rvert\le C_0 \frac{\sum_{k=1}^N e^{\frac{Tk}{N}\left(3r+3\sigma^2\right)}}{\left(\sum_{k=1}^N e^{\frac{Tk}{N}\left(2r+\sigma^2\right)}\right)^{\frac{3}{2}}}$$

Each sum can be rewritten as one expression since it is just a geometric sequence: $$SX=\sum_{k=1}^N e^{\frac{Tk}{N}r}=\frac{(e^{rT}-1)e^{\frac{rT}{N}}}{e^{\frac{rT}{N}}-1}$$ $$SX^2=\sum_{k=1}^N e^{\frac{Tk}{N}(2r+\sigma^2)}=\frac{e^{T(2r+\sigma^2)}-1}{e^{\frac{T(2r+\sigma^2)}{N}}-1}+e^{T(2r+\sigma^2)}-1$$ $$SX^3=\sum_{k=1}^N e^{\frac{Tk}{N}(3r+3\sigma^2)}=\frac{e^{T(3r+3\sigma^2)}-1}{e^{\frac{T(3r+3\sigma^2)}{N}}-1}+e^{T(3r+3\sigma^2)}-1$$

Now, the inequality can be rewritten as $$\left\lvert\frac{\sum_{k=1}^N e^{\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}}-SX}{\sqrt{SX^2}}-\Phi(x)\right\rvert\le C_0 \frac{SX^3}{\left(SX^2\right)^{\frac{3}{2}}}$$

And finally divide by $N$: $$\left\lvert\frac{\frac{1}{N}\sum_{k=1}^N e^{\sigma W_{\frac{Tk}{N}}+\left(r-\frac{\sigma^2}{2}\right)\frac{Tk}{N}}-\frac{SX}{N}}{\frac{\sqrt{SX^2}}{N}}-\Phi(x)\right\rvert\le C_0 \frac{SX^3}{\left(SX^2\right)^{\frac{3}{2}}}$$

The problem here is that $\frac{SX}{N}$ has finite limit as $N\to\infty$, RHS approaches $0$ as $N\to\infty$, but $\frac{\sqrt{SX^2}}{N}\to 0$ as $N\to\infty$. And LHS is infinite.

Where am I wrong?

Thank you in advance.

$\endgroup$
8
  • $\begingroup$ This application of the Berry--Esseen inequality is incorrect -- because that inequality is for sums of indepedendent random summands, whereas your summands are not independent. $\endgroup$ Commented Jan 10 at 21:29
  • $\begingroup$ @IosifPinelis , why are they independent? Wiener process has independent increments: en.wikipedia.org/wiki/… $\endgroup$
    – Paul R
    Commented Jan 10 at 21:30
  • $\begingroup$ Your summands are not increments of the Wiener process. In particular, the random variables $W_{T/N}$ and $W_{2T/N}$ are correlated (with correlation coefficient $1/\sqrt2$) and hence (positively) dependent. $\endgroup$ Commented Jan 10 at 21:38
  • $\begingroup$ Can you explain why? I thought that $W_{\frac{T}{N}}\sim N\left(0,\frac{T}{N}\right)$ and $W_{\frac{2T}{N}}\sim N\left(0,\frac{2T}{N}\right)$. I don't see correlation between them. $\endgroup$
    – Paul R
    Commented Jan 10 at 21:40
  • 1
    $\begingroup$ The correlation, say between $X$ and $Y$, is about the joint distribution of $X$ and $Y$. The individual distributions of $X$ and of $Y$ can tell you nothing about their correlation. $\endgroup$ Commented Jan 10 at 22:49

0

You must log in to answer this question.

Browse other questions tagged .