Miller, chapter 7, End of Chapter questions

[dsake]

Hi David,

could you please expalin how do we get to the answers at the first part of question 3 and question 4?

Many thanks,
dsake

 

[David Harper CFA FRM]

Hi @dsake

Question 3:

Question: 3. You are presented with an investment strategy with a mean return of 20% and a standard deviation of 10%. What is the probability of a negative return if the returns are normally distributed? What if the distribution is symmetrical, but otherwise unknown?

Answer 3. A negative return would be greater than two standard deviations below the mean. For a normal distribution, the probability (one-tailed) is approximately 2.28%. If we do not know the distribution, then, by Chebyshev's inequality, the probability of a negative return could be as high as 12.5% = 1/ 2 × 1/( 22). There could be a 25% probability of a +/– 2 standard deviation event, but we're interested only in the negative tail, so we multiply by ½. We can perform this last step only because we were told the distribution is symmetrical

When the returns are normal (or presumed to be normal), a key action is the standardization of a "raw value,"see https://en.wikipedia.org/wiki/Standard_score

We standardize X with Z = (X - µ)/σ; in this way, the "raw" X value is translated into a standard normal value (quantile). For example, if µ = 10 and σ = 3, then we standardize an X value of 5 with: Z = (5 - 10)/3 = -1.677 and this means that "the value 5 is 1.677 standard deviations below [or to the left, if you like] of the mean of 10." The letter Z is significant because it connotes the standard normal distribution which can be referenced in the Z lookup table, or with excel's command. For example, we can answer the question, "If normal µ = 10 & σ = 3, what is the probability that X will be less than 5" with Pr[Z < (5 - 10)/3] = Pr[Z < -1.667] = NORM.S.DIST(-1.677, true = CDF) = 4.78%

So, above the question is, "What is the probability of a negative return if the returns are normally distributed?" which is:

• Pr [X < 0 ] = Pr[ Z < (0 - 0.20)/0.10 ] = Pr[ Z < -2.0 ]; i.e., the X value zero is 2.0 standard deviations below the mean of zero.

The advantage of the standardized Z is that we know the properties of a standard normal distribution: Pr[Z < 2] = NORM.S.DIST(-2,TRUE) = 2.78%, or we can use the lookup table.

Question 4:

Question 4. Suppose you invest in a product whose returns follow a uniform distribution between − 40% and + 60%. What is the expected return? What is the 95% VaR? The expected shortfall?

Answer 4. The expected return is + 10%. The 95% VaR is 35% (i.e., 5% of the returns are expected to be worse than –35%). The expected shortfall is 37.5% (again the negative is implied).

See https://en.wikipedia.org/wiki/Uniform_distribution_(continuous)

The expected return is the average: (-40% + 60%)/2 = 10%.

The 95% VaR cuts off at the 5% tail. You really don't even need to calculate because -40 runs to 60% in a uniform (flat) way such that the distance is 100%, so we can just add 5%. The 95% ES is the (conditional) average of the 5% tail, but in the uniform, again that's easy because it's just the average of 100% and 95%.

I hope that's helpful!