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Last Minute Doubts & Questions about FRM‐I Exam

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Question 1 – FRM Exam 2008‐F

A portfolio has an average return over the last year of 13.2%. Its benchmark has provided an average return over the same period of 12.3%. The portfolio’s standard deviation is 15.3%, its beta is 1.15, its tracking error volatility is 6.5% and its semi‐standard deviation is 9.4%. Lastly the risk free rate is 4.5%. Calculate the portfolio’s Information Ratio (IR).

A. 0.569B. 0.076C. 0.138D. 0.096

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Answer

C.• Information Ratio = (Average Return on the Portfolio – Average Return on the benchmark)/Tracking Error

Volatility • IR = (13.2 – 12.3)/6.5 = 0.138 • This question tests whether the candidate knows the information ratio. It has a number of destructors that

make it difficult to ‘guess’ the formula• A. Incorrect. • B. Incorrect. • D. Incorrect.

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Question 2

Albert Franklin wants to estimate the market risk premium for the stock of Arlene Pipes Company. He gathers the following information:1. Expected Return for Arlene 10.75%2. Risk free rate 6.00%3. Beta for Arlene 0.60

He believes that stock of Arlene Pipes is fairly priced. Based on above information Market risk Premium for Arlene Pipes Company is closest to:

A. 1.75%B. 14%C. 8%D. 4.75%

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Answer

C (chapter 3, FRM)• Using CAPM Model• 10.75%=6%+0.6(E(RM)‐Rf)• (E(RM)‐Rf)=(10.75%‐6%)/0.6• (E(RM)‐Rf)=7.92%(close to 8%)

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Question 3 – FRM Exam 2009`

An investment bank uses the Exponentially Weighted Moving Average (EWMA) technique with lambda of 0.9 to model the daily volatility of a security. The current estimate of the daily volatility is 1.5%. The closing price of the security is USD 20 yesterday and USD 18 today. Using continuously compounded returns, what is the updated estimate of the volatility?

A. 5.44%B. 3.62%C. 2.96%D. 1.31%

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Answer

B.The current return of the security is = ln (18/20) = ‐10.536%. Using an EWMA model, the updated volatility is given as: V(t) = {lambda* ((V[t‐1]^2) +(1 – lambda)*(current return^2)} ^ 0.5= {0.9 * ((0.015^2) + (1 ‐ 0.9) * ( ‐0.10536^2 )} ^ 0.5= 3.62%• INCORRECT: A – Forgets to square the volatility terms• INCORRECT: C – Forgets to square the volatility terms and to take the square root of the resulting variance,

then miscalculates conversion to percentage.• INCORRECT: D – Forgets to take the square root of the variance, then miscalculates conversion to

percentage.

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Question 4 – FRM Exam 2008

A factor analysis of the dividend‐adjusted returns of ABC Ltd.’s stock price was undertaken to determine which economic factors contributed to its performance. The regression was performed on 460 observations. The results are as follows:

Which one of the following options correctly describes which variables are significant at the 5% level, and the R2 statistic, respectively? Significant Variables at 5% level R2 statistic

A. Intercept; Industrial_index 0.924834B. Intercept; Industrial_index 0.075166C. All_share_index; Industrial_Index 0.924834D. All_share_index; Industrial_Index 0.075166

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Table 1:

Predictor Coeffecient Standard Error of Coefficient

Intercept ‐0.0243 0.005772

All_Share_Index 0.0256 0.017655

Industrial_Index 0.0469 0.006398

Financial_Index 0.0012 0.001412

Table 2

Sum of squared regression (SSR) 12,466,47

Sum of squared Errors (SSE) 1,013.22

Sum of Squared Total (SST) 13,479.69


Answer

A.• The following table shows the test statistics for each of the four variables, calculated by dividing the variable

coefficient by the standard error . The variable is significant if the absolute value of the t‐stat is greater than the critical value from the student’s t‐distribution for 456 degrees of freedom (which is very close to the z‐statistic since the number of observations is so high), i.e. 1.96. Predictor t‐stat Significant?

• Intercept ‐.0243/0.005772 = ‐4.21 Yes • All_share_index .0256/0.017655 = 1.45 No• Industrial_index .0469/0.006398 = 7.33 Yes• Financial_index .0012/0.001412 = 0.85 No• The R2 statistic is defined as the ratio of SSR/SST = 12,466.47/13,479.69 = 0.924834.

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Question 5 – FRM Sample Paper 2011

The following GARCH(1,1) model is used to forecast the daily return variance of an asset:σn2 = 0.000005 + 0.05u2n‐1 + 0.92σ2n‐1Suppose the estimate of the volatility today is 5.0% and the asset return is ‐2.0%. What is the estimate of the long‐run average volatility per day?

A. 1.29%B. 1.73%C. 1.85%D. 1.91%

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Answer

A.• The model corresponds to α = 0.05, β = 0.92, and ω = 0.000005. Because γ = 1− α− β, it follows that γ =

0.03.• Because the long‐run average variance, VL, can be found by VL = ω / γ , it follows that VL = 0.000167. In

other words, the long‐run average volatility per day implied by the model is sqrt(0. 000167) = 1.29%.

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Consider an equity portfolio with market value of USD 100M and a beta of 1.5 with respect to the S&P 500 Index. The current S&P 500 index level is 1000 and each futures contract is for delivery of USD 250 times the index level. Which of the following strategy will reduce the beta of the equity portfolio to 0.8?

A. Long 600 S&P 500 futures contractsB. Short 600 S&P 500 futures contractsC. Long 280 S&P 500 futures contractsD. Short 280 S&P 500 futures contracts

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Answer

D.• To reduce the beta of a portfolio, one needs to lower the position’s exposure to the market. Thus, one needs

to short futures contracts. The number of futures contracts, N, is:

• Thus, one needs to short 280 futures contracts

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contractFuturesoneofSizePositionspotofSizeN oldnew )(

2801000*250

000,000,100)5.18.0(


Question 7‐ FRM Sample Paper 2007

An investor enters into a short position in a gold futures contract at USD 294.20. Each futures contract controls 100 troy ounces. The initial margin is USD 3,200, and the maintenance margin is USD 2,900. At the end of the first day, the futures price drops to USD 286.6. Which of the following is the amount is the variation margin at the end of the first day?

A. 0 B. USD 34 C. USD 334 D. USD 760

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Answer

A.• Since the investor is shorting the futures contract, she/he gains when the futures price goes down. At the

end of the first day, the investor has a gain of (294.2 – 286.6) * 100 = $ 760. Therefore, no variation margin is needed

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Alan bought a futures contract on a commodity on the New York Commodity Exchange on June 1. The futures price was USD 500 per unit and the contract size was 100 units per contract. Alan set up a margin account with initial margin of USD 2,000 per contract and maintenance margin of USD 1,000 per contract. The futures price of the commodity varied as shown below. What was the balance in Alan’s margin account at the end of day on June 5?Day Futures Price (USD)June 1 497.30June 2 492.70June 3 484.20June 4 471.70June 5 468.80

A. USD 1,120B. USD 0C. USD 880D. USD 1,710

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Answer

D.

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Day Future’s Price Daily Gain (Loss) Cumulative Gain (Loss)

Margin Account Balance

Margin Call

June 1 497.30 (270) (270) 1730

June 2 492.70 (460) (730) 1270

June 3 484.20 (850) (1580) 420 1580

June 4 471.70 (1250) (2830) 750 1250

June 5 468.80 (290) (3120) 1710


Question 9

The USD interest rate is 4% per annum and the AUD rate is 6% per annum. Assume that the term structure of interest rates is flat in the US and Australia. Assume current value of AUD to be $0.91. Company ABC, under the terms of a swap agreement, pays 7% per annum in AUD and receives 3% per annum in US$. The principal in the US is 10million USD and that in Australia is 11million AUD. Payments are exchanged each year and the swap will last for 3 more years. Determine the value of swap assuming continuous compounding in all interest rates.

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Solution

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Valuation of currency swap in terms of bonds (millions):

Value of swap in million $ = 11.294*0.91 – 9.7225= $0.5448 million


Question 10 ‐ FRM Sample Paper 2012

A portfolio has a current market value equal to USD 6,247,000 with a daily variance of 0.0002. Assuming there are 250 trading days in a year and that the portfolio return follow a normal distribution. The estimate of annual VaR at 95% confidence level is closes to which of the following?

A. USD 32595B. USD 145,770C. USD 2,297,854D. USD 2,737,868

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Answer

CDaily standard deviation = sqrt ( 0.0002) = 0.01414Annual VaR = 6247000 * sqrt (250) * 0.01414 * 1.645= 2,297,854

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Given the following 30 ordered simulated percentage returns of an asset, calculate the VAR and expected shortfall (both expressed in terms of returns) at a 90% confidence level.‐16, ‐14, ‐10, ‐7, ‐7, ‐5, ‐4, ‐4, ‐4, ‐3, ‐1, ‐1, 0, 0, 0, 1, 2, 2, 4, 6, 7, 8, 9,11, 12, 12, 14, 18, 21, 23

A. VAR (90%) = 10, Expected shortfall = 14B. VAR (90%) = 10, Expected shortfall = 15C. VaR (90%) = 14, Expected shortfall = 15D. VaR (90%) = 18, Expected shortfall = 22

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Answer

B.• Ten percent of the observations will fall at or below the 3rd lowest observation of the 30 listed. Therefore,

the VaR equals 10. The expected shortfall is the mean of the observations exceeding the VaR. Thus, the expected shortfall equals (16 + 14) / 2 = 15

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A portfolio is composed of two securities and has the following characteristics• Investment in security A = USD 1,500,000• Investment in security B = USD 3,000,000• Volatility of security A = 7%• Volatility of security B = 3%• Correlation between security A and B = 10%

What is the closest answer for the portfolio diversified VAR at 95% confident level?

A. USD 7,351B. USD 212,920C. USD 365,715D. USD 234,630

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Answer

D.• To calculate the VAR of the portfolio, we first must calculate the volatility of the portfolio, σ. The variance of

the portfolio equals:• σ2 = (1/3)2 (7%)2 + (2/3)2 (3%)2 + 2*(1/3)*(2/3)*10%*7%*3% = 0.001, Thus, σ = 0.0316• Assuming a 95% confidence interval, VAR = 1.65 * 0.0316 * 4,500,000 = 234,630

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Consider a bond with par value of EUR 1,000 maturity in 3 years, and that pays a coupon of 5% annually.The spot rate curve is as follows:

The value of the bond is closest to:

A. EUR 904B. EUR 924C. EUR 930D. EUR 950

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Term Annual Spot Interest Rate

1 6%

2 7%

3 8%


Answer

B.• Using spot rates, the value of the bond is:• 50/(1.06) + 50/[(1.07)^2] + 1050/[(1.08)^3] = 924.37

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