volatility

sailakshmisuresh

New Member
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

David Harper CFA FRM
Staff member
Subscriber
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?"

Thank you!

David Harper CFA FRM

David Harper CFA FRM
Staff member
Subscriber
@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,