volatility

[sailakshmisuresh]

There was an example in GARP stating "Suppose that an asset price is $60 and that its daily volatility is 2%. This means that a one-standard devaition move in the asset price over one day would be 60*0.02 or 1.20%. If we assume taht the change in the asset price is normally distributed we can be 95% certain that the asset price will be between 60 - 1.96*1.2= $57.65 and 60 + 1.96*1.2 = $62.35."

We have studied that when building confidence intervals the standard error is multiplied with the respective critical value. In the example aforementioned since the standard deviation is 2% shouldn't we divide that by square root of n to get the standard deviation of the sampling distribution (standard error)- more like sampling distribtion of the sample standard deviation). So whatever might be the sample size shouldn't its square root be used to divide 2%. And I am unable to understand why is 2% multiplied by $60. There should be no need to multiply right since 0.02 is the standard deviation that we already found.

Can we write more like
60 +/- 1.96* (0.02/sq rt of n)

Would be grateful if someone can clarify this doubt!
Thanks

 

[David Harper CFA FRM]

HI @sailakshmisuresh I'm glad you shared this (where is it in the GARP curriculum? I looked super quickly in Chapter 3 but I didn't see it ...) because it illustrates a common confusion. Let me write an alternative question:

"Suppose that over the last twenty days (n = 30) we observed an average asset price of $60.00 and the (sample's) standard deviation was $1.20. If the true asset price has a normal distribution (and the daily observations were i.i.d.), what is the 95.0% confidence interval for the true (i.e., population) mean of the asset price?"

Answer: 60.00 ± 1.96 × $1.20/sqrt(30)

Because in my question, the standard deviation is the standard error of the sample mean. We are informed of this the central limit theorem (https://en.wikipedia.org/wiki/Central_limit_theorem) which applies because we are referring to the average (or summation) of a set of i.i.d. random variables. Compare this to your example, which is "merely" descriptive of the single outcome of the random variable; notice how your example resists even calling it a confidence interval which IMO is good form (although I suppose it could ...). In your example, the CI of {57.65, 62.25} is the just an interval for a single outcome of the random variable, we do not want to "re-size" the standard deviation. Both CIs are using a standard deviation, but only the sample mean has a standard deviation (aka, standard error) given by σ/sqrt(n). In regard to 2.0%, that's a units thing, the $60.00 is in dollars so its incongruous to use $ +/- %. I hope that's helpful,

 

[sailakshmisuresh]

This question was mentioned in chapter 14 of “volatility” of GARP Quants book 2019 edition.

Ok so as far as what I understood , you are trying to tell that in the example mentioned we are talking only about a single outcome of a random variable and we are not calculating the population mean ( as per GARP example)

Now in the example that you mentioned shouldn’t we divide by sqrt of 20 ? I didn’t understand as to why $1.2 has been divided by the square root of 30 ? Isn’t the sample size in your example 20?

Thank you!

 

[David Harper CFA FRM]

@sailakshmisuresh in haste I wrote 20 and 30 but fixed it above so that it consistently refers to n = 30.

"Suppose that over the last twenty days (n = 30) we observed an average asset price of $60.00 and the (sample's) standard deviation was $1.20. If the true asset price has a normal distribution (and the daily observations were i.i.d.), what is the 95.0% confidence interval for the true (i.e., population) mean of the asset price?"

Answer: 60.00 ± 1.96 × $1.20/sqrt(30)

It's a sample mean, as the sample mean is a random variable that estimates (aka, as an estimator) the population mean. Each different sample will produce a different sample mean and its this sample mean that has variance of σ/n.

That's for locating it. It fits that the question is in volatility and not where I was looking (Ch 3) because not an statistical inference question, it's just a question about the standard deviation of a distribution. Thanks,