GBM Drift, Volatility, and Expected Stock Prices

# Does GBM stock price model have E[S(t)] unaffected by volatility?


# Does GBM stock price model have E[S(t)] unaffected by volatility?












Many an author claims that, if you model stock prices through GBM, $E[S(t)]=e^{\mu t}$, and the expectation is thus not related to volatility.

I keep running around in circles on this one. First of all, it seems intuitively to have some doubt. But I can argue it either way.

One thing that may affect this is that people are sloppy, I believe, in thinking about the solution to the SDE. the solution is

$$S(t)=S(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t} $$

i.e., a lognormal distribution.

Suppose you were looking at a stock that went up from \$100 to \$105 last year with 20% volatility. It seems many people believe the parameters for the lognormal are thus $\mu=.05$ and $\sigma=.20$ But, it looks to me like the actual parameter that goes in for "mu" for the lognormal is really $.05-\frac{.20^2}{2}$, and to be more accurate it is a smaller number yet since continuous compounding has an impact (i.e., even if $\sigma$ were 0, a number slightly less than $.05$ would be the right rate, $ln(1.05)$ to be exact, so that continuous compounding gives you the 5% one-year return.

So, in that way, it seems like volatility in a GBM reduces returns,since it gets subtracted off.

On the other hand, a lognormal has a mean of $e^{\omega+\frac{\sigma^2}{2}}$, so if $\omega=\mu-\frac{\sigma^2}{2}$ you can convince yourself they do, indeed, cancel out, leaving $e^{\mu t}$. But if this is corect, is it true that the expected value of a price evolving under GBM has no dependence on volatility? If nothing else this seems hard to square with cases where vol is very high, so much so that $\mu-\sigma^2/2$ could become very negative (try $\mu=.10$ and $\sigma=.7$) thus pretty much seeming to guarantee that the $lim$ $t\to \infty$of $S(t)$ goes a.s. to zero.

## Answer by Forgottenscience (score 1, accepted)

https://quant.stackexchange.com/a/42666

Expectations is what we find when we average over all values of uncertainty. If you take a normally distributed variable $X \sim \mathcal{N}(\mu, \sigma^2)$, then it would not be a surprise that the mean is just $\mathbb{E}(X) =\mu$ no matter what $\sigma^2$ is. In this case the underlying model is that $X = \mu + \sigma \eta$ with $\eta \sim \mathcal{N}(0,1)$, we just observe it noisily.

Geometric Brownian motion is equivalently just a noisy version of a first-order ODE

$dS(t) = \mu S(t) dt$,

so it seems reasonable that the expectation would be equal to this simple underlying model of stock prices. This is just $S(t) = e^{\mu t}$ (we have assumed the constant initial price to be 1), which is exactly what you can laboriously derive from the SDE.

## Answer by Kurt G. (score 2)

https://quant.stackexchange.com/a/81158

Let $W$ be a Brownian motion and $X\sim N(\mu t,\sigma^2 t)\,.$ Then \begin{array}{|cc|c|} \hline & S_t & \mathbb E[S_t] & Y_t=\log(S_t) & dY_t & \dfrac{dS_t}{S_t}\\[2mm] \hline (a) & e^{\mu t+\sigma W_t-\sigma^2t/2} & e^{\mu t} & \mu t+\sigma W_t-\tfrac{\sigma^2t}2 & \mu\,dt+\sigma\,dW_t-\tfrac{\sigma^2}2\,dt&\mu \,dt+\sigma\,dW_t\\[2mm] \hline (b) & e^{\mu t+\sigma W_t} & e^{\mu t+\sigma^2t/2} & \mu t+\sigma W_t & \mu\,dt+\sigma\,dW_t&\mu \,dt+\sigma\,dW_t+\frac{\sigma^2}2\,dt\\[2mm] \hline (c) &e^{X-\sigma^2 t/2} & e^{\mu t}\\[2mm] \hline (d) &e^X & e^{\mu t+\sigma^2t/2}\\[2mm] \hline \end{array}

- The last three columns are proved by the Ito formula. I spare us the well-known details.

Proof of the expectations in the second column:

- Proof of (d). Writing $v=\sigma^2 t$ and $\mu'=\mu t$ the expectation is \begin{align} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}e^{\mu'+\sqrt{v}\,x}\,dx\stackrel{complete~the~square}= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(\mu'-\tfrac{(x-\sqrt{v})^2}{2}+\frac v2\right)\,dx \end{align} Can you proceed?

Hint:

> $$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\tfrac{(x-\sqrt{v})^2}{2}\right)\,dx=1\,.$$

- Proof of (d)$\Rightarrow (c)$:

> $$\mathbb E[e^{X-\sigma^2t/2}]=\mathbb E[e^X]\,e^{-\sigma^2t/2}\,.$$Now use (d).

- Proof of (b):

> Set $X=\sigma W_t\,.$ Then use (d).

- Proof of (b)$\Rightarrow (a)$:

> Proceed as in (b) and the proof of (d)$\Rightarrow$ (c).

## Answer by Jan Stuller (score 2)

https://quant.stackexchange.com/a/81181

I like this question for the following reason:

If you believe in the GBM model, then the expected log-returns are indeed affected by the volatility, since:

$$\mathbb{E}\left[\ln\left(\frac{S_t}{S_0}\right)\right]=\mu t - 0.5\sigma^2t$$

However, the regular returns are not affected, since (as already pointed out above) if $X\sim N(\mu, \sigma)$ and $Y:=e^X$ (i.e. $Y$ is log-normal) then:

$$\mathbb{E}\left[Y\right]=e^{\mu+0.5\sigma^2}$$

So in the case of a stock $S_t$, the distribution is lognormal as follows:

$$S_t=S_0e^\left({\mu t-0.5\sigma^2t}+\sigma Z\right)$$

And the $-0.5 \sigma^2t$ ends up cancelling out the $+0.5\sigma^2t$ when we take the expectation.

I think this question is good for the following specific reason: if you want to calibrate the GBM to historical data and you compute (say) the mean $\bar{\mu}$ and the volatility (S.D.) $\bar{\sigma}$ of the log-returns of the S&P500 index, you must realize that the log-returns are distributed with mean $\mu-0.5\sigma^2$ and you must therefore use the volatility estimator $\bar{\sigma}$ to add the term $0.5\bar{\sigma}^2$ back to the mean $\bar{\mu}$: in order to get an actual calibrated estimate of the parameter $\mu$ for the GBM model.

Many people (including practitioners) miss this and they just use $\bar{\mu}$ as an estimator for the GBM $\mu$, but that is wrong.

## Answer by siou0107 (score 1)

https://quant.stackexchange.com/a/81180

A simple observation that may help your intuition: the mode of the lognormal distribution is $e^{m - v^2}$ (here, $m = \mu - \frac{1}{2} \sigma^2$ and $v = \sigma$, the parameters of the underlying Gaussian distribution).

When $v \to + \infty$, this is 0; on the other hand, the $v$ in the denominator of the density makes it near 0 on most points but around the mode. At the limit, all the mass shifts towards 0; the distribution degenerates to a Dirac.

In finance, you usually define lognormal distributions with respect to the log (annualised) mean return $\mu$ rather than the (annualised) mean log return $\mu - \frac{\sigma^2}{2}$, hence $\sigma$ has to be well-defined for the underlying Gaussian distribution to be defined; else, the only thing you can say is "the stock price is positive with mean $S_0 e^{\mu t}$". More on lognormal distributions subtleties here: Why is long term binary put option more expensive than call assuming driftless GBM?

## Answer by mark leeds (score -1)

https://quant.stackexchange.com/a/81152

eSurfSnake and Forgottenscience: I came here to post an "answer" but it's really just a continuation of the topic you are talking about here. I just figured it was best to put this here because it's relevant and my original question was somehow wrongly stated and got closed ? So, thanks for reading this.

Oh, in what follows, assume that returns are small enough so that the log return versus (1+r) difference doesn't come into play. I just want that to not confuse the issue.

First a short summary:

- You solve for the solution to the GBM for dS_t. ( or BM for ds/S ).

- Then, you convert the solution to what that means in terms of expectation in price. It turns out that $E(S_t) = exp(\mu t)$ where $\mu$ is the drift term synonymous with the drift in the original GBM for $dS_t$.

Price is lognormal (even in the GBM case ) but you showed that the solution implies that the expectation of price (in time) is different from what a lognormal (but not in time) rv's expectation usually is, namely:

$$ E(S_t) = S_{0} exp(\mu t) $$

What I mean by this is that, in the world of statistics, where there are no martingales or SDES and time is static, if $S$ is lognormal, then $E(S) = \exp(\mu + \sigma^2/2)$ where $\mu$ and $\sigma^2$ are the parameters of the underlying normal random variable on which exp is being taken.

So, my question is what is the intuition for what is ACTUALLY happening when one imposes a GBM for the price ? Why does the GBM do that to the expectation of price ? I didn't even mention risk neutrality so the solution shouldn't "know about" that theory, should it ? Also, it's not only an issue with the expectation of price.

Consider the infinitesmal log return: We know that

$$ d(log_t) = (\mu - \sigma^2)\times dt + \sigma \times W_t$$

So, this tells me that, infinitesimally speaking, the drift term of the log returns contains an error correction term also. But this then implies that, infinitesimally, $S_t$ itself has to be getting corrected also ( I assumed log and arithmetic are the same ) ?

But then, how can one impose a GBM if it implies that the actual price is being corrected ?

I must be erring somewhere (probably in multiple places ) in my argument ? If someone can explain where my thinking is wrong, it's appreciated. This has been baffling me for the last month or so when I realized just how dependent the strategy I'm working on is on the GBM.

There's some kind of seeming inconsistency that's probably just my wrong thinking tricking me.

No need to be bashful. I'd appreciate hearing that my thinking is really mucked up. That might un-confuse me.

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                                                  Mark
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ADDENDUM: ML: 11/14/2024 #=================================================================

I THINK THAT I HAVE ANOTHER AND BETTER WAY OF ASKING MY QUESTION ABOVE.

Suppose I assume that the returns are a GBM so that $Y = log(S_t)$ is the solution to $dS_t = \mu S_t dt + \sigma S_t W_t$.

I tell you that a stock X has a price of 100.00 dollars at 10:00 am. I then tell you that the expected log return ( i.e: the forecast ) for the stock from 10:00 am to close is zero.

Can one then say that the price of X at close is expected to be 100 dollars ? That's my confusion.