Question: Book 1 - Roy's Safety First ratio

[oatmeals2]

The question is on page 221 of book1:

Portfolio A has a safety-ratio of 1.3 with a thresthold return of 2 percent. What is the shortfall risk for a target return of 2 percent?

I intuitively chose P(Z<1.3)=.903 (wrong)

The correct answer is mentioned by the poster above: 1-N(SFRatio)= 1-.903= 0.0968 (correct)

My question is why are we taking the right tail? To my understanding, P(Z<1.3) represents the probability our returns would be less than threshold(of 2%) while P(Z>1.3) represent our returns above threshhold. I guess I'm going this the wrong way, anyone care to enlighten me?

 

[CFAHouston]

your understanding is correct that we are taking the right tail. However, were not using P(Z<1.3), were using P(Z<-1.3). Remember that SFR is the number of standard deviations below the mean (hence the negative)

So, if you find P(1.3) = .9031. so to find P(-1.3), we subtract .9031 from 1 giving us 0.0968.

said that, i assumes the easiest way to remember is that when it comes to test: choose the highest SF ratio.

Now, if i can just remember which one is SF ratio(threshold), and which one is Sharpe(risk free return)

 

[valuequest]

Here is the explanation which makes sense to me and hopefully to others who consider this exercise:

We do not know the mean, but we do know that the probabilities of achieving a return either above or below the mean are equally 0.5. Conceptually, we know that the 2% threshold falls into the half below the mean. What we want here is the probability of falling below (or further left) this 2% threshold return, or in other words, we want to know the area under the curve which lies below (or further left) of the 2% threshold return. Therefore we need concern ourselves only with the lower (or left) 0.5 of the distribution.

We are given that the SF-ratio=1.3. As CFAHouston said, this tells us the critical fact that the 2% threshold return lies 1.3 standard deviations below (or left) of the mean. Interpreting this value as a z-score and consulting a z-table tells us that 0.4032 lies under the curve between the mean and the 2% threshold. To get the area which falls below (or further left) of this, we simply subtract it from the total lower (or left) half area 0.5. 0.5 - 0.4032 = 0.0968

 

[jamespucyk]

Here's how I think about it. Roy's Safety First and Sharpe are just standardized variables and the variable is standardized and using the Z-table, you will get the chance of achieving a return over the threshold (threshold for Sharpe is RFR). BTW, that is all that a the standardize value will tell you when you translate it with a Z-table.

So, by extension and by converse, the negative standardized variable is just the opposite; the chance or probability of achieving a return lower than the standardized variable. You have to keep in mind, by being given a pop. mean and SD, you are given all you need to equate a value to a probability assuming a normal distribution; which is all that a standardized variable does, allow you to calculate cumulative probabilities.

Basically if you understand how standardized variables relate to Z-table, relate to normal distributions, you understand how this works.