Quiz: Valuation Models

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6 questions to check your understanding of the chapter “Valuation Models”. Choose an answer and select “Check answer” to see the answer and explanation. You can also open “Answer and explanation” to see the answer first.

1. Question 1

   As a hypothetical example, in the Hull-White model, the volatility $\sigma$ of the short rate is 1% and the speed of mean reversion $a$ is 0.2 per year. What is the standard deviation of the short rate 5 years ahead in this model?

   1. A.About 2.24%
   2. B.About 1.78%
   3. C.About 1.47%
   Answer and explanation

   Answer: C

   The standard deviation is about 1.47%. The standard deviation of the short rate $t$ years ahead is $\sigma\sqrt{(1 - e^{-2at})/(2a)}$, and here $2at = 2 \times 0.2 \times 5 = 2$. Since $e^{-2} = 0.1353$, $(1 - 0.1353)/(2 \times 0.2) = 2.1617$, and its square root is 1.470. Multiplying by $\sigma = 1\%$ gives 1.47%.

   2.24% is $\sigma\sqrt{t} = 1\% \times \sqrt{5}$, the value with no mean reversion (the limit as $a$ goes to 0). Because the short rate is pulled back toward its level, the actual spread is smaller than this. 1.78% is the value of $\sigma\sqrt{(1 - e^{-at})/a}$. It drops the 2 in the exponent of the formula. Here $(1 - 0.3679)/0.2 = 3.1606$, its square root is 1.778, and multiplying by $\sigma$ gives 1.78%.

   Learn more: The Hull-White Model

2. Question 2

   As a hypothetical example, in the LIBOR market model (LMM), the volatility of the rate of change of a forward rate is a function of time only: 10% in the first year and 30% in the second. The forward rate is fixed in 2 years. What volatility do you use when you express a caplet on this forward rate with the Black model formula?

   1. A.About 22.36%
   2. B.About 20.00%
   3. C.About 5.00%
   Answer and explanation

   Answer: A

   The volatility is about 22.36%. The volatility in the Black formula is the square root of the time average, over the 2 years, of the square of the LMM volatility. The squared values are $0.10^2 = 0.01$ for the first year and $0.30^2 = 0.09$ for the second, so the time average is $(0.01 \times 1 + 0.09 \times 1)/2 = 0.05$. Its square root is 0.2236.

   20.00% is the simple average of 10% and 30%. Because the squares are averaged, the larger volatility has a stronger effect, and the result is above 20%. 5.00% is the time average of the squares, 0.05, written as a percentage without taking the square root. A volatility is the square root of the average of the squared values, so you must take the square root at the end.

   Learn more: The LIBOR Market Model (LMM)

3. Question 3

   As a hypothetical example, the SABR model has four parameters: $\alpha$, $\beta$, $\rho$ and $\nu$. $\nu$ sets the size of the volatility's own fluctuation. If you increase only $\nu$ and leave the others unchanged, how does the volatility smile of options with one maturity change?

   1. A.The whole curve moves up and the curvature is unchanged
   2. B.The curvature gets stronger and the center hardly moves
   3. C.The curvature gets weaker and the center hardly moves
   Answer and explanation

   Answer: B

   Increasing $\nu$ strengthens the curvature of the smile and raises the implied volatility at both ends. The value at the center (near the forward) hardly moves.

   The curvature gets weaker when $\nu$ is lowered. The whole curve moves up when $\alpha$ is raised. The slope gets steeper when $\rho$ is made more negative or when $\beta$ is lowered.

   Learn more: The SABR Model

4. Question 4

   In the local volatility model, when the local volatility is a function of the underlying price only, what smile movement does the model predict when the underlying price rises?

   1. A.The smile moves toward higher strikes
   2. B.The smile does not move, and the values at each strike stay
   3. C.The smile moves toward lower strikes
   Answer and explanation

   Answer: C

   The smile the model predicts moves toward lower strikes. When the forward moves from $f_0$ to $f$, the smile after the move can be written, keeping only the first term of the expansion, as $\sigma_0(K + f - f_0)$ using the smile before the move $\sigma_0$, where $K$ is the strike. When the underlying price rises, the value at a given strike $K$ becomes the value at a higher strike before the move, so the smile appears to shift toward lower strikes.

   In the market, the underlying price and the smile move in the same direction, so the model's prediction is the opposite of the market. "The smile does not move, and the values at each strike stay" is the view that the smile stays tied to the strike (sticky strike). In this model, the local volatility is determined by the underlying price alone, so when the price moves, the position of the smile shifts too. The view that the smile does not move contradicts the model's prediction.

   Learn more: Local Volatility Models

5. Question 5

   As a hypothetical example, in the Heston model, the speed of mean reversion $\kappa$ of the variance is 3, the long-run variance $\bar v$ is 0.04, and the volatility of variance $\xi$ is 0.4. Which describes the movement of the variance $v_t$ correctly?

   1. A.It meets the Feller condition and stays positive without reaching 0
   2. B.It meets the Feller condition and touches 0 repeatedly
   3. C.It fails the Feller condition and stays positive without reaching 0
   4. D.It fails the Feller condition and touches 0 repeatedly
   Answer and explanation

   Answer: A

   The Feller condition is met, and the variance does not reach 0 but stays strictly positive. The condition is $2\kappa\bar v \ge \xi^2$. The left side is $2 \times 3 \times 0.04 = 0.24$ and the right side is $0.4^2 = 0.16$, so the condition is met.

   When the condition is met, the variance does not reach 0. Touching 0 repeatedly is the behavior when the condition is not met. If you forget the factor 2 on the left side and compare $\kappa\bar v = 0.12$ with $\xi^2 = 0.16$, you wrongly conclude that the condition is not met. Even when the condition is not met, the variance only touches 0 and returns, and never stays at 0.

   Learn more: The Heston Model

6. Question 6

   As a hypothetical example, in a one-factor Gaussian copula, the expected loss of the equity tranche from 0 to 3% is 80%, and that from 0 to 7% is 50%. Each is valued with the base correlation of its detachment point, and expected losses are shares of each tranche's notional. What is the expected loss of the 3–7% tranche?

   1. A.15.7%
   2. B.27.5%
   3. C.30%
   Answer and explanation

   Answer: B

   The expected loss is 27.5%. The expected loss of the tranche from $A$ to $B$ can be written as the difference between the expected losses of the tranches from 0 to $B$ and from 0 to $A$. $E[L_{0,B}]$ is the expected loss of the tranche from 0 to $B$, as a share of its notional. Writing $A$ and $B$ as numbers in % (3 and 7), the formula is $(B \times E[L_{0,B}] - A \times E[L_{0,A}])/(B - A)$. The numerator is $7 \times 0.5 - 3 \times 0.8 = 1.1$. The denominator is $7 - 3 = 4$. Dividing gives $1.1/4 = 0.275$ (27.5%).

   15.7% is $1.1/7$, which divides by $B = 7$ instead of $B - A = 4$. The 3–7% tranche is 4% wide, so you divide by that width. 30% is the difference between 80% and 50% used as it is, which ignores the difference in the width of the ranges.

   Learn more: Copulas for Credit