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yuxdhk
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### Estimating Volatility is easier than Estimating Stock Price?

One of professor said that at one of the MIT open course, is this true? If yes, why?

Alan
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### Re: Estimating Volatility is easier than Estimating Stock Price?

It's true -- assuming of course that the professor meant a future stock price.

A way to see it is to consider estimating the drift $\mu$ and volatility $\sigma$ in the stock price model $dX_t = \mu dt + \sigma dB_t$, where $X_t = \log P_t$ and $B_t$ is a Brownian motion. Here $P_t$ is the stock price.

I will just give hints.

Suppose you have $N$ observations of historical prices over $(0,T)$, where $N \Delta t = T$. To be clear, the setup is that the (log)-prices are discrete observations from a complete path sample, $\{X_t: 0 \le t \le T\}$, generated by the model.

What are the standard estimators $(\hat{\mu}_N,\hat{\sigma}_N)$ for $(\mu,\sigma)$, also known as maximum likelihood estimators, given the observation series $\{\Delta X_1, \Delta X_2, \cdots, \Delta X_N\}$?

How do they improve (or not) as $N$ grows large and $T$ is fixed? You should convince yourself that $\hat{\mu}_N = (\log P_T/P_0)/T$ for *all* $N$ and so does *not* become more precise as $N \rightarrow \infty$, whereas $\hat{\sigma}_N \rightarrow \sigma$ in the same limit.

In other words, even given nanosecond observations of the $X_t$, you cannot improve your knowledge of $\mu$ with $T$ fixed. The implication is that, if you use your estimates from data over $(0,T)$ to make forward predictions for $P_{2 T}$ and the volatility over $(T, 2 T)$, the latter can be made with arbitrary accuracy while the former remains intrinsically uncertain.

A slightly different take on the matter is that, even if $(\mu,\sigma)$ were known precisely, it's still going to be more difficult to estimate $P_{2 T}$ than the volatility over $(T, 2 T)$  -- given that the predictions are being made from time $t = T$.  It's a similar idea, but I leave a more careful statement of it to you.