# Q.320 Quantitative Analysis

A random variable X with an exponential distribution has the following probability density function: FX(x) = &lambda;e- x&lambda;You are required to determine the probability density function of the random variable X, given that Y = X2 A 1 – e-&lambda;y1/2 B – e-&lambda;y1/2 C 1/2 &lambda;y- ½ e- &lambda;y1/2 D 1 The correct answer is: C From first principles, it can be shown that:FY(y) = P(Y &le; y) = P(X2 &le; y) = P(X &le; y1/2) $$P[X\le { y }^{ \frac { 1 }{ 2 } }]=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ \lambda { e }^{ -\lambda x }dx }$$ From calculus, $$\int { { e }^{ -cx }dx=-\frac { 1 }{ c } } { e }^{ -cx }\quad$$Thus, $$P[X\le { y }^{ 1/2 }]=\int _{ 0 }^{ { y }^{ 1/2 } }{ \lambda { e }^{ -\lambda x }dx } ={ \left| \frac { -\lambda }{ \lambda } { e }^{ -\lambda x } \right| }_{ 0 }^{ { y }^{ 1/2 } }\\ =-{ e }^{ -\lambda { y }^{ 1/2 } }--{ e }^{ 0 }\\ =1-{ e }^{ -\lambda { y }^{ 1/2 } }\\ P[X\le { y }^{ 1/2 }]={ F }_{ Y }(y)=1-{ e }^{ -\lambda { y }^{ 1/2 } }\\ { f }_{ Y }(y)={ F }'_{ Y }(y)=0-\left( -\frac { 1 }{ 2 } \lambda { y }^{ -1/2 }{ e }^{ -\lambda { y }^{ 1/2 } } \right) \\ =\frac { 1 }{ 2 } \lambda { y }^{ -1/2 }{ e }^{ -\lambda { y }^{ 1/2 } }$$ *User Question: I could not understand the problem and its solution at all. Can you please explain it from scratch?

FRM Part 1

# Q.948 Valuation and Risk Models

An analyst is comparing the STDEV or GARCH methodology with that of the RiskMetric&reg; approach for estimating VaR using historical data. He wrote down the following similarities between both methods. Which of the following similarities is incorrect? A Both methods belong to the parametric class of risk assessing models B Both methods attempt to estimate conditional volatility C Both methods apply equal weights to all the periods D Both methods use recent historic data for assessing risk The correct answer is: C The standard deviation models STDEV or GARCH apply equal weights to all the windows of past data, while the RiskMetric&reg; approach applies higher weights on more recent data. The weights decline exponentially to zero as returns become older. *User Question: Does GARCH apply equal weights to all past period? I think not

FRM Part 1

# Q.886 Financial Markets and Products

## 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? ?