P1.T2.70. Standard error Page 101 of Question set

[Dr. Jayanthi Sankaran]

70.3 Assume the population of a hedge fund returns has an unknown distribution with mean of 8% and volatility of 10%. From a sample of 40 funds, what is the probability the sample mean return will exceed 10%?

(a) 10.0%
(b) 10.3%
(c) 10.7%
(d) 11.1%

Answer is (b) 10.3%

The Z statistic turns out to be 1.2658 ~ 1.27 such that Probability (Z >= 1.27) = 0.5 - 0.3980 = 10.2%. However, you have rounded the Z statistic to 1.26 instead, as a result of which Prob (Z>=1.26) = 0.5 - 0.3962 = 10.38% ~ 10.4%
Although, by itself, it is trivial, for examination purposes, how would one decide between answers (a) 10% and
(b) 10.3%

Thanks!
Jayanthi

 

[brian.field]

The exam would not (hopefully) have an ambiguous question.

 

[ShaktiRathore]

Hi
Yes Jayanthi Brian is right exam shall not have ambigous question,the calculated answer whether u round the tstat or not will alwaud lie nrarer to the correct option. For exam always go for the answer most nearer to the calculaed value u got after approximations or not. There might be only one value that shall be the correct answer and that shall lie near to your calculated value, there wiuld be few one or two such questions in the exam otherwise all answers shall directly point to the correct option. Plrese pay attention to the details of the question and be careful in calculations so as not to choose incorrect option in a hurry.
Thanks

 

[Dr. Jayanthi Sankaran]

Thanks, Shakti and Brian for that vote of confidence!

Jayanthi

 

[Sahil1999]

70.3 Assume the population of a hedge fund returns has an unknown distribution with mean of 8% and volatility of 10%. From a sample of 40 funds, what is the probability the sample mean return will exceed 10%?

(a) 10.0%
(b) 10.3%
(c) 10.7%
(d) 11.1%

Answer is (b) 10.3%

The Z statistic turns out to be 1.2658 ~ 1.27 such that Probability (Z >= 1.27) = 0.5 - 0.3980 = 10.2%. However, you have rounded the Z statistic to 1.26 instead, as a result of which Prob (Z>=1.26) = 0.5 - 0.3962 = 10.38% ~ 10.4%
Although, by itself, it is trivial, for examination purposes, how would one decide between answers (a) 10% and
(b) 10.3%

Thanks!
Jayanthi

Hi David (@David Harper CFA FRM ),
In 70.3, I'm unable to understand why are we dividing by the Standard Error and not the volatility itself. Request your support!

Kind Regards,
Sahil

 

[David Harper CFA FRM]

Hi @Sahil1999 Because it's asking about a sample mean, so central limit theorem (CLT) applies.

Each fund has μ = 8.0% with σ =10.0%.
For a single fund, the Pr(R > 10%) = 1 - Pr[R < (10% - 8.0%)/10%)] = 1 - Pr (Z ≤ 0.20) = 1 - 57.93% = 42.07%.

But if we ask about the mean of a sample of 40 such funds (although as noted above this question is badly worded, sorry, it is a variation on a GARP source that was not well worded) then we need to utilize CLT such that the standard deviation of the sample mean has a standard deviation given by σ/sqrt(n), and in this way:

the Pr(R_avg > 10%) = 1 - Pr[R_avg < (10% - 8.0%)/(10%/sqrt(40))] = 1 - Pr (Z ≤ 1.265) = 1 - 89.70% = 10.30%. Hope that's helpful!

P.S. see my post here https://forum.bionicturtle.com/threads/what-is-the-standard-error-of-the-sample-mean.5605/post-15828

 

[Sahil1999]

Hi David (@David Harper CFA FRM ),

This helped me a great deal in understanding the concept better. Since, we are using a distribution of samples of 40 funds, we are looking at Standard Error as a metric to standardise the deviation to get the statistic, right ?
We can not use volatility since, it's representative of only one fund.

P.S. the post for understanding standard error was extremely helpful.

Kind Regards,
Sahil

 

[David Harper CFA FRM]

Hi Sahil (@Sahil1999 ) Glad it helped! I think you are close but I would rephrase as follows: "because we are measuring the dispersion of a sample mean (i.e., a sample of 40 funds), we look at the standard error which is a standard deviation but rather than for a single fund, it's a standard deviation of a sampling distribution."

In case it's further helpful, here is an internal slack entry I wrote for my team last year to describe the standard error ...

Here a customer has today asked a profound question, why did you use the standard deviation and not the standard error? The standard error lies at the heart of statistics (and machine learning, in fact), but so many of my customers cannot define it. When I can, I use dice. Dice are relatable. Even better is one six-sided die. What is the standard deviation (aka, dispersion) of a such a single die? The most useful variance formula is E(X^2) - E(X)^2; hence I often link candidates to refresher at https://en.wikipedia.org/wiki/Variance. A six-sided die is a random space {1, 2, ..., 5, 6} and the square of that vector is {1, 4, ..., 25, 36}. The average of the first set is 3.5, and the average of the second set is ~15.167. Applying the formula, the variance = 15.167 - 3.5^2 = 2.92 and the square root of 2.92 is 1.70. To a new learner, this is often a satisfying exercise. To identify dispersion as a feature of a die.

So what's a standard error? Now imagine that I hold the single die behind my back (at a table behind my back). You can't see the die. You only know it is random, you don't know how many faces nor even if it's fair! Yikes, you do not know the true population distribution. Instead, I roll the die behind my back and report the results of ten rolls to you: {1, 5, 3, 4, 1, 1, 4, 5, 1, 5}. The average of this particular sample happens to be 3 and the average of the squared vector is 12. Applying the same formula, the variance E(X^2) - E(X)^2 = 12 - 3^2 = 3.0 such that the standard error is SQRT(3) = 1.73.

This shows how the standard error is a standard deviation (sweet, yes? I wish I had a nickel for every time I had to remind learners that a standard error is already a standard deviation). What is the difference? The standard error was computed from the sample. If I roll ten times again, I'll get a different standard error. My random number generator was a six-sided die, so I'm not surprised to see it's nearby. I expect it to be inequal but nearby. Why did I make the bold claim that standard error is common in machine learning? Because in real-life, we never get the population parameters; we never really know the true die's look. All of these big data algorithms are generally computing statistics on samples and they compute their associated standard errors. I hope that's interesting.

... and based on your observation, yesterday I wrote this (in our internal slack) about how I would today re-write my old "lazy" question. Hopefully my two re-writes of this question are informative:

An example of a lazy (ie., very old but it's my own) question that insufficiently helps new learner:

70.3 Assume the population of a hedge fund returns has an unknown distribution with mean of 8% and volatility of 10%. From a sample of 40 funds, what is the probability the sample mean return will exceed 10%?

CLT for sample mean is important, highly discussed, and easily confused. Especially the related question of whether we can use the normal approximation if the sample is large/small and whether we know/don't know the distribution.If I were to write it today, I'd write something like this:

70.3. The distribution of a hedge fund's monthly return is unknown to us but somehow we do know this monthly return has a mean of 8.0% with volatility of 10.0%. Over 40 months when the returns are independent (i.i.d.), what is the probability that the average arithmetic mean (aka, sample mean) over this 3 1/3 year period exceeds 10.0%?

or if I wanted to preserve the academic vibe of the original, something like:

70.3. Let's assume the periodic return of a hedge fund is a random variable drawn from a population where the mean is 8.0% per period with volatility of 10.0% but the distribution is unknown to us. From a sample of 40 funds, what is the probability the sample mean return will exceed 10%?

 

[Sahil1999]

Hi Sahil (@Sahil1999 ) Glad it helped! I think you are close but I would rephrase as follows: "because we are measuring the dispersion of a sample mean (i.e., a sample of 40 funds), we look at the standard error which is a standard deviation but rather than for a single fund, it's a standard deviation of a sampling distribution."

In case it's further helpful, here is an internal slack entry I wrote for my team last year to describe the standard error ...

... and based on your observation, yesterday I wrote this (in our internal slack) about how I would today re-write my old "lazy" question. Hopefully my two re-writes of this question are informative:

The tagged example of the dice makes it absolutely clear David (@David Harper CFA FRM). Thank you so much for the comprehensive explanations.

Have a great day ahead

Kind Regards,
Sahil