# Q.3621 Quantitative Analysis

In a game, a coin is flipped. If the coin is heads, the player rolls one die. If the coin turns up tails, the player rolls two dice and the player moves their playing piece that number of spots shown on the die or dice. Given that on a player's turn, he moves 5 spaces, what is the probability he flipped tails on the coin? A $${1}/{10}$$ B $${1}/{5}$$ C $${2}/{5}$$ D $${1}/{3}$$ The correct answer is: C $$P\left( 5 \right) =P\left( { 5 }/{ T } \right) \ast P\left( T \right) +P\left( { 5 }/{ H } \right) \ast P\left( H \right)$$ $$P\left( 5 \right) =\left( { 1 }/{ 9 } \right) \ast \left( { 1 }/{ 2 } \right) +\left( { 1 }/{ 6 } \right) \ast \left( { 1 }/{ 2 } \right) = {5}/{36}$$ $$P\left( { T }/{ 5 } \right) =P\left( { 5 }/{ T } \right) \ast { P\left( T \right) }/{ P\left( 5 \right) }$$ $$P\left( { T }/{ 5 } \right) =\left( { 1 }/{ 9 } \right) \ast { \left( { 1 }/{ 2 } \right) }/{ \left( { 5 }/{ 36 } \right) }=\bf { 2 }/{ 5 }$$ *User Question: How do you get P(5/H)=1/6??? Please give detailed explanation

FRM Part 1

# Q.3682 Equity Valuation: Concepts and Basic Tools

An equity analyst is comparing the price multiples of two companies – Quartic Ltd. and Recon Inc. While examining the balance sheet of the two companies, the analyst finds out that Quartic capitalizes its advertisement costs while Recon expenses out its advertisement costs. Assuming that both companies are similar in all other aspects, Quartic Ltd. is most likely to have: A a higher price-to-earnings ratio than Recon Inc. B a lower price-to-earnings ratio than Recon Inc. C the same price-to-earnings ratio as Recon Inc. The correct answer is: B) As Quartic capitalizes its advertisement costs, it is most likely to report higher earnings as compared to company Recon. Hence, Recon Inc. is most likely to have a higher P/E ratio. *User Question: Comments and answer are not in sync

CFA Level 1

# Q.3037 Market Risk Measurement and Management

You have been hired on the trading floor and one of the traders comes over and asks about the impact of a price change on her VaR made of a long position in stock A, whose value stood at 100 as of yesterday. Assume you are using a 95% confidence historical VaR (based on 260 days moving window of historical data). Further, assume that the 16 worst 1-day returns of stock as of yesterday were as follows: -9.5, -8, -7.6, -7.4, - 7.2, -7.18, -7.1, -6.9, -6.57, -6.56, -6.45, -6.4, -6.25 -6.05, -5.99,-5.85. Assume the price of the stock increased by 10% between yesterday and today. What will the value of today's 95% VaR (in absolute value)? A &dollar;6.25 B &dollar;6.875 C &dollar;10 D Cannot conclude The correct answer is: B Today's stock price is $100 &times; (1 + 10%) =$110 Given that the 95% worst return is -6.25%, the new 95% VaR will be $110 -$110 &times; (1 - 6.25/100) = -&dollar;6.875 *User Question: The 13th observation (-6,25%) is in the 5% above 95% (0,05*260=13). The 14th observation is actually the highest loss within 95% confidence intervall ( -0,0605) If we don't know when exactly the worst 16 observations were recorded, we don't know if maybe one of them fell out of the sample when we include the newest return...

FRM Part 1

# Q.3728 Common Probability Distributions

A vehicle repairs assembly has a total of 100 jerks and other repair work machines in constant use. The probability of a machine breaking down during a given day is 0.004. There are days when none of the machines break down. However, during some days, one, two, three, four, or more machines break down. What is the probability that fewer than 3 machines break down during a particular day? A 0.007726 B 0.6698 C 0.9923 The correct answer is: C)The number of breakdowns takes on a binomial distribution with n = 100 and &theta; = 0.004Fewer thanù implies 0, or 1, or 2 machines break downTherefore, P(fewer than 3) = P(0 breakdowns) + P(1 breakdowns) + P(2 breakdowns)= 100C0 * 0.0040 * 0.996100 + 100C1 * 0.0041 * 0.99699 + 100C2 * 0.0042 * 0.99698= 0.6698 + 0.2690 + 0.05347= 0.9923 *User Question: What's the value of 100C2, and how do you get it?

CFA Level 1

# Q.3012 Market Risk Measurement and Management

## Q.3593 Market Organization and Structure

An investment bank offers its customers the option to carry out leveraged trades. The investors are required to maintain a margin of 30% and pay a commission of 0.25% of the trade value. An investor acquires 2,000 shares each at a price of $30. If the shares are currently trading at &dollar;40 and the borrowing cost is 8%, then the return generated by the leveraged trade is closest to: A 89.75% B 9.08% C 90.75% The correct answer is: A) Total fund required to acquire 2,000 shares = 2,000 &ast; &dollar;30 = &dollar;60,000 Margin required for the trade =$60,000 * 30% = $18,000 Commission =$60,000 * 0.25% = $150 Total out-of-pocket investment required for the trade =$18,000 + $150 =$18,150 Total funds required = &dollar;60,000 + &dollar;150 = &dollar;60,150 Funds borrowed = &dollar;60,150 - &dollar;18,150 = &dollar;42,000 Interest cost = &dollar;42,000 &ast; 8% = &dollar;3,360 Profit earned in the leveraged trade = 2,000&ast;(&dollar;40-&dollar;30) - Commissions - Interest paid = 2,000&ast;&dollar;10 - &dollar;60,000&ast;0.25% - &dollar;80,000&ast;0.25% - &dollar;3360 = &dollar;16,290 ROE = &dollar;16,290 / (&dollar;18,150) = 89.75% *User Question: There are two commissions one when you buy and the other when you sell. (you have to subtract them both) ?

## Q.332 Quantitative Analysis

Two random variables X and Y are such that V[X] = 4V[Y] and Cov[X,Y] = V[Y] Let E = X + Y and F = X – YFind Cov[E, F] A V[Y] – V[X] B Cov[X,Y] C V[Y] D 3V[Y] The correct answer is: D Cov[E,F] = Cov[X + Y,X – Y] = Cov[X,X] – Cov[X,Y] + Cov[Y,X] – Cov[Y,Y] = V[X] – V[Y] = 4V[Y] – V[Y] = 3V[Y] Logic applied: I. Given a random variable X, the covariance between X and itself is simply its varianceII. Cov[X,Y] = Cov[Y,X] *User Question: Why is Cov[X + Y,X – Y] = Cov[X,X] – Cov[X,Y] + Cov[Y,X] – Cov[Y,Y] ?

## Q.509 Quantitative Analysis

Which of the following is not a characteristic describing the dynamic nature of a white noise process? A The unconditional mean and variance must be constant for any covariance stationary process B The absence of any correlation means that all autocovariances and autocorrelations are not zero beyond displacement zero C Events in a white noise process do not exhibit any correlation between the past and the present D Both conditional and unconditional means and variances are the same for an independent white noise process The correct answer is: B The lack of any correlation means that all autocovariances and autocorrelations are zero beyond displacement zero. Displacement is the distance covered by a moving body from a central point. *User Question: Can anyone explain what beyond zero' imply? ?