# Q.687 Financial Markets and Products

CFA Level 1

## Q.4620 Valuation and Risk Models

The futures price of an asset is USD 40, and the annual volatility of the futures price is 20%. If the risk-free rate is 5%, what is the value of a put option to sell futures in 6 months for USD 45? A USD 0.028 B USD 4.498 C USD 0.026 D USD 5.520 The correct answer is: D In this case, $${\text{F}}_{0}$$=40, K=45, r=0.05, s=0.20, T=0.5 The following formula gives the value of the put option: $${ \text{P} }_{ 0 }=\text{K}{ \text{e} }^{ -{ \text{r} }\text{T} }\times \text{N}\left( -{ { \text{d} } }_{ 2 } \right) -{ \text{S} }_{ 0 }{ \text{e} }^{ -{ \text{r} }\text{T} }\times \text{N}\left( -{ { \text{d} } }_{ 1 } \right)$$ Where: $${\text{P}}_{0}$$= value of the put option $${\text{F}}_{0}$$= current futures price K= strike price s= volatility of the futures price r= risk-free rate T= time $${ \text{d} }_{ 1 }=\cfrac { \text{ln}\left( \cfrac { { \text{F} }_{ 0 } }{ \text{K} } \right) +\cfrac { { \sigma }^{ 2 }\text{T} }{ 2 } }{ \sigma \sqrt { \text{T} } } =\cfrac {\text{ln}\cfrac { 40 }{ 45 } + \cfrac{{ 0.20 }^{ 2 }}{2}\times 0.5 }{ 0.20\sqrt { 0.5 } } =-0.76214 \\ { \text{d} }_{ 2 }={ \text{d} }_{ 1 }-{ \sigma \sqrt { \text{T} } }=-0.9036 \\ \text{N}\left( -{ \text{d} }_{ 1 } \right) =\text{N}\left( 0.762 \right) =0.7764 \\ \text{N}\left( -{ \text{d} }_{ 2 } \right) =\text{N}\left( 0.9036 \right) =0.8159$$ The value of the put option is given by: $${ \text{P} }_{ 0 }=\text{K}{ \text{e} }^{ -{ \text{r} }\text{T} }\times \text{N}\left( -{ { \text{d} } }_{ 2 } \right) -{ \text{S} }_{ 0 }{ \text{e} }^{ -{ \text{r} }\text{T} }\times \text{N}\left( -{ { \text{d} } }_{ 1 } \right) \\ { P }_{ 0 }=45{ \text{e} }^{ -0.05\times 0.5 }\times 0.8159 -40{ \text{e} }^{ -0.05\times 0.5 }\times 0.7764=\text{ USD } 5.520$$ *User Question: I do not understand why the calculation of d1 does not take into account the risk free rate. Any thoughts as to why that is the case? My d1 = -0.5856 .

## Q.378 Quantitative Analysis

A random sample of 50 FRM exam candidates was found to have an average IQ of 125. The standard deviation among candidates is known (approximately 20). Assuming that IQs follow a normal distribution, carry out a statistical test (5% significance level) to determine whether the average IQ of FRM candidates is greater than 120. Compute the test statistic and give a conclusion. A Test statistic: 1.768; Reject H0 B Test statistic: 2.828; Reject H0 C Test statistic: 1.768; Fail to reject H0 D Test statistic: 1.0606; Fail to reject H0 The correct answer is: A The first step: Formulate H0 and H1H0: &mu; = 120H1:&mu; &gt; 120Note that this is a one-sided test because H1 explores a change in one direction onlyUnder H0, (x&#772; - 120)/(&sigma;/&radic;n)&nbsp;&#8767;N(0,1) Next, compute the test statistic: = (125 &ndash; 120)/(20/&radic;50) = 1.768Next, we can confirm that P(Z > 1.6449) = 0.05, which means our critical value is the upper 5% point of the normal distribution i.e. 1.6449. Since 1.768 is greater than 1.6449, it lies in the rejection region. As such, we have sufficient evidence to reject H0 and conclude that the average IQ of FRM candidates is indeed greater than 120. Alternatively, we could go the “p-value way” P(Z > 1.768) = 1 – P(Z &lt; 1.768) = 1 – 0.96147 = 0.03853 or 3.853%This probability is less than 5% meaning that we have sufficient evidence against H0. This approach leads to a similar conclusion. *User Question: Can you please explain how did we get P(Z > 1.6449) = 0.05 ?

## Q.4666 Valuation and Risk Models

The standard deviation of the daily portfolio value changes is given as 0.0231, and its mean as 0.0012. Given that there are 250 trading days in a year, what is the annual 99% VaR for the portfolio according to the delta-normal model? A 0.78 B 0.83 C 0.58 D 0.63 The correct answer is: B According to delta-normal mode, VaR is given by: $$\text{VaR}={ \mu }_{ \text{P} }+{ \sigma }_{ \text{P} } { \text{U} }$$ Since we are dealing with 99% VaR, it implies that $$\text{U}=-2.326$$ So that daily VaR is given by: \begin{align*} \text{VaR}=& 0.0012+0.0231\times-2.326=-0.05253 \\ \text{Annual VaR}=& \sqrt{250}\times-0.05253=-0.8306 \end{align*} *User Question: While scaling up and finding an annual VAR, shouldn't the return be multiplied by 250 and the standard deviation be multiplied by square root of 250. Since we will use the property that the sum of 'n' IID with mean R and Std Dev D is n*R and (square root of n)*(Std Dev D). This will give us a normal distribution with mean of n*R and std dev of (square root of n*Std Dev D), which will be 250*0.0012= 0.3 (mean) and standard deviation of (square root of 250)*0.0231= 0.3652. Now applying U = -2.33 we get VAR as 0.5509. How can we scale up a value of a normal distribution at a particular percentile directly by the square root of time as it is done in the solution?

## Q.528 Quantitative Analysis

An autoregressive process is considered stationary if: A The roots of the characteristic equation lie on the unit circle B The roots of the characteristic equation lie outside the unit circle C The roots of the characteristic equation lie inside the unit circle D The characteristic equation are of order 1 The correct answer is: B In any autoregressive process, the roots of the characteristic equation must lie outside the unit circle, which means the absolute value of the roots must be larger than one. *User Question: which roots and unit circle are these? what do they imply. ?

## Q.3190 Risk Management and Investment Management

Jack Marconi is an equity strategist at Gandhara Investment and is evaluating the performance of four large-cap equity portfolios: Azgard, Lambda, Tricky, and Jackpot. As part of his analysis, Jack computes the Sharpe ratio and the Treynor measure for all four funds. Based on his finding, the ranks assigned to the four funds are as follows: $$\begin{array}{|c|c|c|} \hline Fund & Treynor\quad Measure\quad Rank & Sharpe\quad Ratio\quad Rank \\ \hline Azgard & 1 & 4 \\ \hline Lambda & 2 & 3 \\ \hline Tricky & 3 & 2 \\ \hline Jackpot & 4 & 1 \\ \hline \end{array}$$ The difference in rankings for Funds Azgard and Jackpot is most likely due to: A Different benchmarks used to evaluate each fund's performance B A difference in risk premiums C A lack of diversification in Azgard Fund as compared to Jackpot Fund D None of the above The correct answer is: C The most likely reason for a difference in ranking is due to the absence of diversification in Azgard Fund. The Sharpe ratio measures excess return per unit of total risk, while the Treynor ratio measures excess return per unit of systematic risk. Since Azgard Fund performed well on the Treynor measure and so poorly on the Sharpe measure, it seems that the fund carries a greater amount of unsystematic risk, meaning it is not well diversified and unsystematic risk is not the relevant risk measure. *User Question: Hello, I believe the explanation is not correct (the answer is correct though). Azgard is the fund that performed poorly on the TREYNOR ratio and performed well on the Sharpe (not vice versa as you state it) Therefore, Azgard carries a greater amount of systematic risk (Beta is greater ,therefore Traenor is low) or it was not well diversified. Please comment because the current explanation is confusing - e.g. if a porftolio performes well on the Treynor it means it is WELL diversified.