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yuxdhk
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Estimating Volatility is easier than Estimating Stock Price?

May 31st, 2019, 6:19 am

One of professor said that at one of the MIT open course, is this true? If yes, why?
 
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Alan
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Re: Estimating Volatility is easier than Estimating Stock Price?

May 31st, 2019, 3:29 pm

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. 
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