Garp sample exam 1, Level 1, question 1

[sarita]

Dear David, i need help with another Garp sample sample question:

Asume that a random variable follows a normal distribution with a mean of 50 and a standard deviation of 10. What percentage of this distribution is between 55 and 65?

1) 4.56%
2) 8.96%
3) 18.15%
4) 24.17%

May i bother you to explian the response to this question?

I understand that upper/lower range = x+/- zvalue*sigma. and if you 55 and 65, you will get 1.5 sigma and .5 sigma. However, i don't undertand how number 4 is the answer.

Many thanks in advance.
S

 

[David Harper CFA FRM]

Hi Saray,

N(1.5) = NORM.S.DIST(1.5) = 93.3913%; i.e., ~93% of the area under the standard normal is to the LEFT of +1.5 standard deviations
N(0.5) = NORM.S.DIST(0.5) = 69.1462%; i.e., ~69% of the area under the standard normal is to the LEFT of +0.5 standard deviations
... NORM.S.DIST() is excel only b/c i am too lazy to lookup same in the table in the book.
Note: you cannot answer this question as given without a lookup table!

so to get the area between, we have to subtract 69% from 93% = 24.173% is the "slice" of the area between 0.5 and 1.5

David

 

[cidare]

Hi David,
There is another question from the same exam paper.

17. A risk manager for bank XYZ, Mark is considering writing a 6 month American put option on a non-dividend paying stock ABC. The current stock price of the option is USD 50 and the strike price of the option is USD 52. In order to find the no-arbitrage price of the option, Mark uses a 2-step binomial tree model. The stock price can go up or down by 20% each period. Mark’s view is that stock price has an 80% probability of going up each period and a 20% probability of going down. The risk free rate is 12% per annum with continuous compounding.
What is the risk-neutral probability of the stock price going up in a single step?
a. 34.5%
b. 57.6%
c.65.5%
d. 80%

The correct answer is b.

In my opinion, according to the binomial tree method, the calculation should be as follows:
P(up) = (exp(r*delta t) - d)/(u-d) = [(exp(0.12*3/12)-exp(-volatility*sqrt(delta t))]/[exp(volatility*sqrt(delta t))-exp(-volatility*sqrt(delta t))] = (1.0305-exp(-0.2*sqrt(3/12)))/(exp(0.2*sqrt(3/12))-exp(-0.2*sqrt(3/12))) = (1.0305-0.9048)/(1.1052-0.9048)=0.6270 = 62.7% .

Is it possible that there exist some miscalculations?
I would like your help.
Many thanks in advance.

 

[David Harper CFA FRM]

Hi cidare,

Isn't the flaw in your approach: you don't have volatility from the question? Your formula looks correct but where do you get the 0.20 for the volatility input?
... for example, if you were to infer it from u = 1.2, then u = exp(sigma*sqrt[t]) --> sigma = ln(u)/sqrt(t) = ~ 36%.

HOWEVER, i already cited the question as problematic, to GARP, for its imprecision and i think your approach is related to my own problems with the question. Technically, i have three problems with the wording, including "up or down by 20% per period." By the answer, they obviously mean by this that u = 1.2 and 0.8, but I think it is imprecise (it does not say per STEP, "period" in this context is arguably ambiguous).

Your approach while flawed--correct me if i am wrong please--by assuming a volatility is understandable as you are employing the "matching volatility with u and d."

I blame the imprecision of the question, where the question should either use exact terms and/or, as i try do in my questions, qualify with "(u = 1.2)" ... Nothing is gained by omitting u = 1.2

hope that helps, David

 

[cidare]

I have figured out what I misunderstood. I am very grateful.
Thank you David.