Confidence interval formula - CFA L1 Quant

Hello

Can someone explain why we don’t use the confidence interval formula (point estimate ± (reliability factor × standard error)) here and instead we multiply the reliability factor by the standard deviation directly?

You are using standard error.

For the return for one year, the standard error (of the mean) is:

Standard\ error = \frac{\sigma}{\sqrt{n}} = \frac{\sigma}{\sqrt{1}} = \frac{\sigma}{1} = \sigma
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How does n=1? Question gives the mean return of the period 1926-1997, so n should be equal to 1997 minus 1926 which equals 71.

You’re calculating a confidence interval for the return on small stocks next year. One year.

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For whatever it’s worth, if you have data from 1926 to 1997, the number of years is 72, not 71.

(Count them on your fingers.)

Pardon my impudence but just to elucidate I make an attempt below … and it may totally differ from other protagonists. To some extent it is even unorthodox and lame. And I am not going to use any formula or work out the problem for you. You are most certainly capable of working out the math:

  1. Your quest and use of std. error comes from sampling error. Sampling error is inevitable when you are dealing with sample. A population has no std. error because it cannot have sampling error. All values are deemed true for all point estimates in a population.

  2. Hence you should treat std. error as the rate at which the sample std. Dev. approaches pop. std. Dev. It is not exactly linear (root n) but it may help if you think that way. Hence as the sample size increases the std. error will start approximating the Sx to Pop. Std. Dev.

  3. Looking at this way, inevitably the pop. std. error (for want of a better word) must be lowest of all…it is the pop. std. dev(as given)/(N)^0.5

  4. Just one more thing remains… if in qs. sample std. dev. given- use that. if in qs. pop std. dev. given-use that. if given both , of course use pop std. dev.

And in continuation, I hope you know when to use z- value and when to use Student’s t- distribution. Both are standard normal variate or as as you call it reliability factor.

Bill gave the (correct) answer, but I’ll try and reason why his answer is indeed correct.
This requires a brief review of Probability Theory.

Recall from Probability Theory that, if X and Y are independent normal random variables with
mean \mu_{X} and \mu_{Y}, and variances \sigma_{X}^2 and \sigma_{Y}^2 , respectively,
then the sum of X and Y is also normal with mean \mu_X + \mu_Y and variance \sigma_{X}^2 + \sigma_{Y}^2.

More generally, if you have N independent normal random variables X_{1}, X_{2}, ... , X_{N} ,
with means \mu_{X_{1}}, \mu_{X_{2}}, ... , \mu_{X_{N}} and variances \sigma_{X_{1}}^2, \sigma_{X_{2}}^2, ..., \sigma_{X_{N}}^2 ,
then their sum is also normal with mean \mu_{X_{1}} + \mu_{X_{2}} + ... + \mu_{X_{N}}
and variance \sigma_{X_{1}}^2 + \sigma_{X_{2}}^2 + ... + \sigma_{X_{N}}^2 .

This can be easily proven through straight-up integration or more elegantly through MGF, but this is beyond the scope for CFA exams. One important thing to note is that you sum the variances, NOT the standard deviations (don’t make this mistake in your exam!)

For this problem, you are given that the average return is .177, so \mu = .177 and the unbiased sample standard deviation is .339, so \sigma = .339
(we assume here that the unbiased volatility is equal to the population volatility). If we treat the return of each individual year as an independent normal random variable, then they each follow a normal distribution with mean \mu = .177 and volatility \sigma = .339 .

Suppose you have N random variables X_1, X_2, ..., X_N , all having the same \mu = .177 and same standard deviation \sigma = .339 .
From Probability Theory above, we know that the sum S = X_1 + X_2 + ... + X_N is normal
with mean .177*N and variance .339^2*N .

A 95% confidence interval for the mean of S is
.177N ± 1.95996*(.339^2*N)^.5
= .177
N ± 1.95996*.339*N^.5

However, you want the confidence interval for the average (and not S),
so you divide this quantity by N.
Let this average be represented by a random variable X. Then,

A 95% confidence interval for the mean of X is
(N*.177)/N ± (1.95996*.339*N^.5)/N
= .177 ± (1.95996*.339*N^.5*N^.5)/(N*N^.5)
= .177 ± (1.95996*.339*N)/(N*N^.5)
= .177 ± (1.95996*.339)/(N^.5)
which agrees with the formula provided by Bill (with the value \sigma = .339 plugged-in)

The problem asks for the confidence interval for one year.
That means N = 1, resulting in the confidence interval ( -48.74%, 84.14%)

You’ll notice that as N increases, the confidence interval becomes more narrow.
For example, when N = 2, the confidence interval is (-29.28%, 64.68%).
This makes sense: this shows that as your portfolio becomes more diversified, the portfolio risk decreases. It’s instructive to note that as N approaches infinity, the confidence interval reduces to the single number, the mean. At this point, your portfolio is a risk-free asset with return equal to .177.
You have diversified away all risk in your portfolio.

For reference, if you instead used the T-distribution, the information comes from 72 years worth of data. That means there are 71 degrees of freedom and the critical value for a 95% confidence interval and 71 degrees of freedom is approximately 1.994 (compare this with the critical value of 1.95996 for a Z-distribution). If you use an online calculator, it gives the critical value for T as 1.993943. The confidence interval when N = 1 would then be .177 ± 1.994*.339 or (-49.90%, 85.3%).
The interval is slightly wider because T has a heavier tail than the Z-distribution.

That happens sometimes. Usually by accident. Don’t let it throw you.

The formula
(point estimate ± (reliability factor × standard error))
Is used when you are using a sample to predict the mean of the population

That is not what you are asked here. You are using in an historical data to predict future data.

The questions should not ask about “confidence interval” but interval or range.
This questions does not really relate to what is discussed in chapter 10.