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Aim: to show that changing a probability measure via the application of a Radon-Nikodym derivative preserves variance of a log-normally distributed random variable (for the case when variance is non-stochastic, i.e. constant in $\omega$): including discrete random variables that converge to log-normal in the limit.

1. Continous Case:

I go with the well-known Geometric-Brownian-Motion model:

$$X_T:=X_0+\int_{h=0}^{h=T}\mu X_hdh+\int_{h=0}^{h=T}\sigma X_hdW_h$$

Above, $\mu$ and $\sigma$ are both constant, and the solution to the SDE is:

$$ln\left(\frac{X_T}{X_0}\right)=\mu T-0.5\sigma^2T+\sigma W_T$$

Which means:

\begin{equation} \boxed{ln\left(\frac{X_T}{X_0}\right)\sim N\left(\mu T-0.5\sigma^2T,\sigma \sqrt{T}\right)} (1) \end{equation}

If we want to eliminate or change the drift $\mu$ in the original SDE, we can work with the solution to the SDE above as follows:

$$ln\left(\frac{X_T}{X_0}\right)=\mu T-0.5\sigma^2T+\sigma W_T=\\=\mu T-0.5\sigma^2T+\sigma W_T+rT-rT=\\=-0.5\sigma^2T+\sigma\left(W_T+\frac{\mu-r}{\sigma}T\right)+rT$$

Defining $\hat{\mu}:=\frac{\mu-r}{\sigma}$, we can apply Girsanov Theorem to $W_T+\hat{\mu}T$ above as follows:

$$\frac{\mathbb{dQ}}{\mathbb{dP}}=exp\left(-W_T\hat{\mu}+0.5\hat{\mu}^2 T\right)$$

To get:

$$ln\left(\frac{X_T}{X_0}\right)=r T-0.5\sigma^2T+\sigma W_T^{\mathbb{Q}}$$

If we set $r=0$, the drift term would be eliminated.

Conclusion: We see that applying the Radon-Nikodym derivative has changed the drift, but preserved the variance of the variable $X_T$.

2. Discrete Case:

We define $X_n$ as follows:

$$X_n=X_0u^kd^{n-k}$$

Above, $u$ and $d$ are "up" and "down" multiplicative factors on a binomial tree, we therefore have $k\sim Bin(n,p)$.

Going with the Cox-Ross-Rubinstein parameters for "up" and "down", we set $u:=e^{\eta\frac{T}{n}+\sqrt{\frac{T}{n}}\sigma}$ and $d:=e^{\eta\frac{T}{n}-\sqrt{\frac{T}{n}}\sigma}$.

First of all, using CLT (which works well for Binomial if $p$ is close to 0.5), we have:

$$\lim_{n\to\infty}k\xrightarrow{d}N(np,\sqrt{np(1-p)})$$

We can write:

$$ln \left( \frac{X_n}{X_0} \right) = k \left( 2 \sqrt{\frac{T}{n}} \sigma \right) + n\left(\eta \frac{T}{n}-\sqrt{\frac{T}{n}}\sigma\right) $$

Which gives:

$$\mathbb{E}\left[ln \left( \frac{X_n}{X_0} \right)\right]=\sqrt{Tn}\sigma(2p-1)+\eta T $$

$$Var\left(ln \left( \frac{X_n}{X_0} \right) \right)=4T\sigma^2p(1-p))$$

And again using CLT, we get:

\begin{equation} \boxed{\lim_{n\to\infty}\left( \frac{X_n}{X_0} \right)\xrightarrow{d}N(\sqrt{Tn}\sigma(2p-1)+\eta T,2\sqrt{T}\sigma\sqrt{p(1-p))}} (2) \end{equation} Specifically, if we take $p=0.5$, we get the beatiful result:

\begin{equation} \boxed{\lim_{n\to\infty}ln\left(\frac{X_n}{X_0}\right)\xrightarrow{d} N\left(\eta T,\sigma \sqrt{T}\right)} (3) \end{equation}

And trivially, setting $\eta:=\mu -0.5\sigma^2$ will mean that the discrete model converges to the continuous GBM.

Now I would like to show that applying Radon-Nikodym derivative to the discrete case will preserve the variance, at least in the limit.

We know that $k\sim Bin(n,p)$, therefore we can write $k:=\sum_{i=1}^{n}B_i$, with $B_i\sim Bernoulli(p)$. We could then define the Radon-Nikodym for the discrete case as follows: $\frac{\mathbb{dQ}}{\mathbb{dP}}=a$ when $B_i=1$ and $\frac{\mathbb{dQ}}{\mathbb{dP}}=b$ when $B_i=0$ for some constants $a$ and $b$. Then we have:

$$\mathbb{E}^{\mathbb{Q}}[k]=\mathbb{E}^{\mathbb{P}}\left[\frac{\mathbb{dQ}}{\mathbb{dP}}k\right]=\sum_{i=0}^{i=n}p*a*1=npa=np^{\mathbb{Q}}$$

This brings us back to equation (2), where we can replace $p$ with $p^{\mathbb{Q}}$. Under the assumption that the original $p$ was set to 0.5, it is clear that $p^{Q}$ is no longer 0.5 (if for example $a:=\frac{2}{3}$ and $b:=\frac{4}{3}$, we have $p^{\mathbb{Q}}=\frac{1}{3}$).

It is now clear that after changing the measure, the variance is no longer preserved.

QUESTION: The fact that the discrete case converges to the GBM continuous case when we set $p=0.5$, but doesn't seem to converge perfectly onto the GBM when $p\neq0.5$ strikes me as odd. I would also like to get an intuition for why the variance is preserved under change of measure for the continuous case, but isn't for the discrete case.

I would appreciate any tips or hints, and specifically, references to any papers that might deal with this specific problem. Thank you so much,

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    $\begingroup$ I am following this with interest from the other site. One thought I had: in a binomial tree, you cannot change the drift by more than a certain amount (because the expected value of the drift adjusted process cannot be higher than the maximum possible value or lower than the minimum). But, if indeed you choose a drift adjustment like that, which attains the maximum or minimum value, then the variance must vanish. So basically you can’t always preserve variance if the range of outcomes is finite. $\endgroup$
    – dm63
    Jan 29, 2022 at 20:09
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    $\begingroup$ math.stackexchange.com/questions/79875/… This seems to be the answer. Need to preserve quadratic variation otherwise you have null sets not the same in the two measures $\endgroup$
    – dm63
    Jan 30, 2022 at 0:40
  • $\begingroup$ You wrote $\mathbb{E}^{\mathbb{Q}}[k] = \mathbb{E}^{\mathbb{P}}[\frac{d\mathbb{P}}{d\mathbb{Q}} k]$. It should be the other way around. $\endgroup$ Jan 30, 2022 at 11:51
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    $\begingroup$ Sorry I only had time to look at this now. If I find the time I'll try to contribute in the next week, but no guarantees 😅. $\endgroup$ Jan 31, 2022 at 19:04
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    $\begingroup$ @ShannonStarr: thank you for the feedback, I corrected the typo with the Radon-Nikodym derivative. I had originally posted a related question on MathSE here, but it received no satisfactory answer, whilst a number of contributors reached out to me to say the problem was of interest. I then spent a couple of days on the problem to try to solve it: the post here on MathOverflow shows where I am currently stuck. I think the questions is borderline professional mathematics. $\endgroup$ Feb 1, 2022 at 10:54

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