Financial Risk Manager

Early Bird Episode No. 2 - Question 1

2008-01-18T18:25:37-08:00

QUESTION:

See image below for illustration of four-step binomial tree.

Let’s assume a four-step binomial tree for the random walk of a stock that starts at price S(0). At each node, the stock can jump up (u) or down (d). This is a recombining binomial tree. That means, for example, the middle node (ud) at the second step is reached in two ways, either: an up jump (u) followed by a down jump (d), or a down jump (d) followed by an up jump (d). In other words, ud = du. Similarly, the node (uud) at the third step can be reached three ways: uud = udu = duu.

Assume S(0) = $10, p = 50%, u = 1.1 and d = 0.9.

That is, today’s stock price, S(0), is $10. The probability (p) of an up jump is 50%. The magnitude of an up jump (u) is 1.1 and the magnitude of a down jump (d) is 0.9. For example, at the first step, the price of S(1) will be either $11.00 ($10 x 1.1 = $11, in the case of an up jump) or $9.00 ($10 x 0.9 = $9, in the case of a down jump).

(i) How many total paths are possible through the tree (separately counting even those paths that redundantly reach the same final nodes)?

(ii) How many paths can be taken to reach the middle node at S(4). That is, the node reached with two ups and two down jumps (e.g., uudd)? (Please try to express as a combination)

(iii) At the end of the tree are five possible events. Which of the five is most likely and what is its probability?

(iv) What is the stock’s expected value at S(4)?

(v) The final distribution at S(4) is a discrete binomial distribution, what is the variance of this binomial distribution?

(vi) Tough (unfair) question: assume we are not given p, u, or d. Rather, assume that each step is one year, that the annualized volatilty of the stock is 20%, that the (annual) riskless rate is 5%, and that each step is three months; i.e., S(4) is one year. What are p, u, and d?

(vii) Tougher still: Now assume the annualized volatility is 20% and the (annual) riskless rate is 5%, but instead assume each time step is three months (0.25) such that S(4) is reached in one year. What are p, u, and d?

ANSWER:

Here is a link to an EditGrid XLS that computes the answers below.

(i) 2^4 = 16 possible paths.

(ii) This is given by a combination, the same combination that is embedded in the binomial probability formula. You can visually count the number of paths to reach node (UUDD) is six. This is given by combination for “2 taken out of 4 objects” or 4C2. That’s because this node has two “success” (or up jumps) out of four total. 4C2 = 6.

(iii) The most likely occurrence is the mean event, which is S(4) = UUDD. The probability of this event is given by the binomial distribution, where the number of up jumps (U) is two. In other words, the most likely is “two successes out of four trials” and P(X=2) = 6/16 = 37.5%. It is useful to recall we can straightaway use the binomial distribution: P(X=2) = (4C2)(50%^2)(50%^2) = 37.5%.

(iv) The stock’s expected value is given by $10(1.1)(1.1)(0.9)(0.9) = about $9.80. Note that because this is a recombining tree, the associative property of multiplication holds, and the order does not matter; e.g., same result if $10(0.9)(0.9)(1.1)(1.1).

It is sort of okay to give $10 as the expected value instead of $9.80. If you understand the difference! $10 would be based on an arithmetic mean, while $9.80 is geometric. The geometric return will always be lower because volatility erodes returns!

(v) The variance of a binomial distribution is n(p)(1-p) or (n)(p)(q) where q = (1-p). In this case, variance = (4)(50%)(50%) = 1.0. That means, of course, that the standard deviation is also 1.0 as SQRT(1) = 1.

(vi) See John Hull page 252: under this approach we can “match up and down to volatility.” Up (u) = EXP(20%) = 1.221, Down (d) = EXP(-20%) = 0.819 and probability of up (p) = (a -d)/(u-d) where a = EXP(riskless rate). Here, a = 1.051 so p = 0.577.

(vii) This scales the steps. Up (u) = 1.105, Down (d) = 0.905, a = 1.013 so that p = 0.538

Early Bird Episode No. 8 - Question 3

2008-03-01T18:09:22-08:00

Question:

Assume Hewlett-Packard’s (HPQ) stock price is $50 with an annualized volatility of 40%. The riskless rate is 4%. For simplicity’s sake, assume HPQ does not pay a dividend (actually, they do). Consider a one year European call option with a strike of $50. To summarize:

Stock (S) = $50
Strike (K) = $50
Volatility (sigma) = 40%
Term (T) = 1.0
Dividend = 0
Riskless rate (r) = 4%

To perform this calculation, you will need the standard normal cumulative distribution.
Click here to view a lookup table.
You can also get N(z) in Excel with the function =NORMSDIST(z)

(i) What is the option’s delta?
(ii) Use this delta in a sentence (i.e., interpret its meaning)
(iii) If we assume (as we often do) that the stock price follows geometric Brownian motion (GBM), how are the stock’s PERIODIC RETURNS assumed to be distributed?
(iv) Under GBM again, how are the future stock PRICE LEVELS assumed to be distributed?
(v) What does the Black-Scholes option pricing model give for the price of the call option?

Answer:
(i)
A Eupean call option on a non-dividend paying stock has a delta = N(d1)
d1 = 0.3 (see spreadsheet here for detail)
N(d1) = NORMSDIST(0.3) = 0.62

(ii)
Possibilities include,
“if the stock price changes by X%, the call option price changes by 0.62X%”
“0.62 is the (instantaneous) rate of change of the option price with respect to the stock price”
“0.62 is the slope of the tangent line on the plot of stock price (x) to option price (y)”
“0.62 is the slope of the linear approximation given by the first-order derivative with respect to stock price”

(iii)
GBM assumption: periodic returns are normally distributed

(iv)
GBM assumption: future price levels are lognormally distributed
Please make sure you understand why:
the continously compounded periodic return is given by LN(ST/S0).
In other words, LN(ST/S0) is normally distributed.
That means, by definition, that ST/S0 is lognormally distributed (i.e., both the future price level and the future price as a ratio of the current price are lognormally distributed)

(v)
d1 = 0.3
N(d1) = 0.618
d2 = -0.1
N(d2) = 0.46
Black-Scholes = (S)N(d1) - (K)EXP[(-r)(T)]N(d2)
= (50)(0.618) - (50)(0.9608)(0.46) = $8.79 (approximately)[/img][/img]

Early Bird Episode No. 5 - Question 1

2008-02-04T15:27:49-08:00

Question:

As of Feb 2008, the spot price of gold is $900. The August 2008 gold futures price is $921 (+ 6 months) and the Feb 2009 gold futures price is $923 (+ 12 months). Finally, the current riskless rate is 3%.

In summary:

Gold spot = $900
Gold futures (+ 6 months) = $921
Gold futures (+ 12 months) = $923
Riskless rate = 3%

Let’s assume continuous compounding/discounting. (Please note: all numbers are significantly rounded from actual, for convenience’s sake).

(i) If storage costs are 6% per annum, what does the cost of carry model predict for the Feb 2009 futures price (+12 months)?
(ii) If we assume gold has no convenience yield (nor financing costs above the riskless rate), what does the six month contract imply is the per annum storage cost of gold?
(iii) If storage costs are $2 per ounce per year, rather than a proportional cost, what does the cost of carry model predict for the Feb 2009 futures price (+12 months)?
(iv) What is the delta of six month futures contract (including the storage cost above)? Please try and show this by taking the first derivative!
(v) Bonus: (i) Is this gold futures curve in contango? (ii) is this gold futures curve in normal contango?

Answer:

(i)
Where u = proportional storage costs, the COC model says:
F=(S)EXP([r+u]T). In this case, F = (900)EXP([3%+6%][1]) = $984.76

(ii)
Where u = proportional storage costs, the COC model says: F = (S)EXP([r+u]T).
So: F/S = EXP([r+u]T)
Taking natural log of both sides: LN(F/S) = [r+u]T
Such that: u = LN(F/S)/T - r
In this case, u = LN(921/900)/0.5 - 3% = 1.613% per annum

(iii)
Lump-sum storage costs = U = ($2)EXP([-3%][1]) =$1.94
F = (S + U)EXP[(rate)(T)] = (900 + 1.94)EXP[(3%)(1)] = $929.41

(iv)
Since F = (S)EXP[rT], the first derivative dF/dS = EXP[rT]
On the six-month contract (T=1), delta = EXP[(3% + 1.613%)(0.5)] = 1.023
Or, if you did not include storage costs, delta = EXP[(3%)(0.5)] = 1.015

(v)
Contango = Forward/futures price > Spot Price. The gold futures curve is indeed “in contango”
Normal contango = Forward/futures price > Expected future spot price. Unknown. Neither normal contango nor normal backwardation can be currently observed.
For more on the difference between (normal) contango and backwardation, see this post.

Early Bird slide 7

2008-06-27T10:22:16-08:00

David,

On slide 7 you show a rule.

y= 10+3x -x^3

y’= 3+3x^2

I have 2 questions

so the ‘ of a constant times X is the constant?

why does -x^3 become positive

Early Bird Episode No. 5 - Question 3

2008-02-05T20:08:03-08:00

Question:

Assume we have fitted a few nonlinear functions to explain a company’s annual sales over time (as a function of the year T). In these functions, sales are given as a function of the year (T):

1. S(T) = 2-(T^2)+(T^3)
2. S(T) = (10)EXP[T^0.25]
3. S(T) = LN[6*T^2]
4. Bonus: What is the relative growth rate of function 2: S(T) = (10)EXP[T^0.25]?

For each of the equation, evaluate the first derivative at the fifth year (T=5); i.e., the derivative of S with respect to T.

Answer:

1.
S(T) = 2-(T^2)+(T^3)
S’(T) = 0 - (2)(T) + (3)(T^2)
S’(5) = (2)(5) + (3)(5^2) = 85

2.
S(T) = (10)EXP[T^0.25]
S’(T) = (10)EXP[T^0.25]*(1/4)(T)(-3/4)
S’(5) = (10)EXP[5^0.25]*(1/4)(5)(-3/4) = 0.65

3.
S(T) = LN[6*T^2]
S’(T) = (1/6[T^2])(12)(T) = 2/T
S’(T) = 2/5 = 0.40

4.
S(T) = (10)EXP[T^0.25]. Taking natural log of both sides:
LN[S(T)] = LN(10) + [T^0.25]
LN[S(T)] = constant + [T^0.25]
Relative growth rate is equal to differential of this natural log function:
Growth rate (T) = (0.25)[T^(-0.75)]
Growth rate (5) = (0.25)[5^(-0.75)] = 7.48%

Early Bird Episode No. 3 - Question 3

2008-01-23T22:31:16-08:00

Question:

We regressed the monthly returns of two HFRI Hedge Fund Indices (actual data as of Jan 2008, not hypothetical): the Equity Hedge Index (E) and the Distressed Securities Index (D). We will call the Equity Hedge Index (E) the independent variable; and we will call Distressed Securities Index (D) the dependent variable.

Here were the results of the analysis for the five-year period ending Dec 2008:

n = 60 (sample size of 60 monthly returns, or 5 years)
Covariance (E,D) = 0.00013 (or 0.013%)
Variance (E, independent) = 0.00026 or 0.026%
Variance (D, dependent) = 0.00014 or 0.014%
y-intercept (D-intercept) = 0.0072 or 0.72% (if y=mx + b is the linear regression, then b = 0.72%)

(i) What is the slope of the regression line. That is, what is (m) in the equation y = mx + b (or D = mE + b)?
(ii) If the independent variable (E) equals 5%, what does the regression line predict for the dependent variable (D)?
(iii) What is the correlation coefficient between E and D?
(iv) What is the coefficient of determination (r^2) for the regression line?
(v) Bonus: What is another way to calculate the coefficient of determination if you are given, for example, that both SSE and SSR equal 0.004?
(vi) Bonus: Is the correlation coefficient (r) significant?
(vii) Gold star bonus (bionic turtle tough!): If the sum of squared errors (SSE) is 0.004, what is a 95% confidence interval that bounds the projected dependent variable (D) if E equals 5%?

Answer:

You can analyze the EditGrid spreadsheet here.

(i) Slope = Covariance(E,D)/Variance(E) = 0.013%/0.026% = 0.5 (note this is rounded up from actual of 0.4986)

(ii) y = (slope)(X) + intercept. In this case, E = (slope)(D) = (0.5)(5%) + 0.72% = 3.22%

(iii) Correlation = Covariance (E,D) / [(Standard Dev E)(Standard Dev D)]. In this case, = 0.013%/[[SQRT(0.026%)][SQRT(0.014%)]] = about 68.1% (rounded from actual of 69.47%)

(iv) Coefficient of determination = the square of the correlation coefficient. So, it equals about 46.4%%

(v) r^2 = SSR/SST or 1 - (SSE/SST). Since SSR + SSE = SST, SST = 0.004 + 0.004 = 0.008. And r^2 = 0.004/0.008 = 0.5 (rounded from the actual of 0.4825)

(vi) The test for significant of the correlation coefficient (r) given sample size (n) is given by (r)*SQRT[n-2]/SQRT[1-r^2]. In this case, the test statistic = (68%)SQRT(60-2)/SQRT[1-(68%^2)] = about 7 (or up to 7.35 depending on how exact). As that is significantly greater than 2.0 (the “sweet spot"), we can conclude the correlation coefficient is indeed significant.

(vii) First, we need the standard error of the estimate (SEE) which can be figured from the SSE: SEE = SQRT(SSE/(n-2)). In this case, SEE = SQRT(0.004/58) = 0.008305. As the SEE is analogous to the standard deviation, we can use it to bound the expected value at E = 5%:

Lower: D (L) = mx + b - [SEE][NORMSINV(97.5%)], and
Upper: D (U) = mx + b + [SEE][NORMSINV(97.5%)]

In this case,

Lower: D (L) = 3.22% (the predicted D) - [0.008305][1.96], and
Upper: D (U) = 3.22% (the predicted D) + [0.008305][1.96]

Depending on rounding, that produces a 95% confidence interval of 1.59% and 4.85%.

Early Bird Episode No. 3 - Question 2

2008-01-23T17:15:02-08:00

Question:

Next year, the economy will be in one of three states: low growth (30% likely), moderate growth (60% likely) and high growth (10% likely). Assume both Google’s (G) and Yahoo’s (Y) sales are a function of the economic state, such that:

If economic growth is low (30% likely), Google (G) will grow 20% and Yahoo (Y) will grow 4%
If economic growth is moderate (60% likely), Google will grow 30% and Yahoo will grow 4%
If economic growth is high (10% likely), Google will grow 60% and Yahoo will grow 10%

(i) What is Google’s expected growth?
(ii) What is the covariance between Google (G) and Yahoo (Y)?
(iii) What is Google’s variance?
(iv) What is the correlation (coefficient) between Google and Yahoo?

Answer:

See the EditGrid worksheet for a complete solution.

(i) Google’s expected growth is the weighted average: (30% low)(20%) + (60% moderate)(30%) + (10% high)(60%) = 30%

(ii) COVAR(G,Y) = E[GY] - E[G]E[Y]. Since the expected product of G*Y = 1.56%, COVAR(G,Y) = 1.56% - (30%)(4.6%) = 0.0156 - 0.0138 = 0.0018

(iii) VARIANCE(G) = COVAR(G,G) = E[G^2] - [E(G)]^2.

Since the set of G = (20%, 30%, 60%), the set of G^2 = (4%, 9%, 36%). Therefore, the E[G^2] = (30%)(4%)+(60%)(9%)+(10%)(36%) = 10.2%. And E(G) = 30%. So, variance is given by:

VARIANCE (G) = E[G^2] - [E(G)]^2 = 10.2% - (30%)^2 = 10.2% - 0% = 1.1%
(Note: the standard deviation is therefore about 10.954%)

(iv) Correlation = COV/(STD)(STD). In this case, (0.0018)/[(0.109545)(0.018)] = 91.29%.

Early Bird Episode No. 2 - Question 3

2008-01-19T14:35:04-08:00

Question:

Assume mutual fund period returns are normally distributed with mean expected return of 8% and standard deviation of 10%.

(i) If a particular return is +15%, how many standard units is that from the mean?

(ii) Approximately what percentage of funds should return more than +15%? (simplistic single period assumption, nothing fancy)

(iii) What is the 95% parametric value at risk (VaR), in return terms?

(iv) Actual equity returns tend to violate normality. Cite at least one key difference between the empirical tendencies of equity returns and normality?

Answer:

(i) If a particular return is +15%, how many standard units is that from the mean?

To express a value in STANDARD UNITS, we divide it’s distance from the mean by the standard deviation.
In this case, (15% - 8%)/10% = +0.7 standard units

(ii) Approximately what percentage of funds should return more than +15%?

This is the area “to the right” of the +0.7 standard units, in the standard normal distribution.
The percentage of funds that return less than or equal to 15% is given by the cumulative probability distribution function.
In this case, P (X <= 0.7 standard units) = NORMSDIST(0.7) = NORMDIST(15%, 8%, 10%, TRUE) = 75.8%. That is the probability of less than or equal to +15%. Therefore, the probability of greater than +15% = 1 - 75.8% = 24.2%.

(iii) What is the 95% parametric value at risk (VaR), in return terms?

You could answer two ways.
In all cases, NORMSINV(5%) = -1.645, such that with 95% confidence:
Relative VaR = (10%)(-1.654) = -16.5%, and/or
Absolute VaR = 8% + (10%)(-1.654) = 8.45%

(iv) Actual equity returns tend to violate normality. Cite at least one key difference between the empirical tendencies of equity returns and normality?

As Linda Allen says (Chapter 2), asset returns tend to be:

1. Fat-tailed (kurtosis > 3, leptokurtosis)
2. Skewed
3. Unstable (time-varying; the parameters change over time!)

For reflection:

You need to be able to STANDARDIZE quickly.

Note how simple is relative VaR. (At least this approach. We can get more complicated, but the starting point is this ‘delta normal’ approach. Specifically, this approach is PARAMETRIC, as opposed to historical simulation or Monte Carlo. And WITHIN PARAMETRIC, this uses a normal distribution as opposed to some other distribution). I like to say it is “scaling volatility” because all we do is multiply volatility by the critical value that is a function of our confidence.

Note that 95% VaR confidence is a ONE-TAILED test that gives us a critical value of (-)1.645. If you are new to FRM/risk, you’ll soon see that we have two really common critical values: 1.645 @ 95%/5% and 2.33 @ 99%/1%. (95% confidence is another way of saying 5% significance).

Please FOCUS on Linda Allen’s THREE points. We do all this model work with the distributions. But we can’t forget that actual, empirical asset returns do not conform (they are not “well behaved"). Notice this is almost comical as collectively, these “problems” are deadly. The ‘unstable’ is particularly vexing: our parameters can be correct today and incorrect tomorrow.

Early Bird Episode No. 2 - Question 2

2008-01-19T13:51:56-08:00

Question:

A company produces widgets but with a 2% defect rate (i.e., two widgets out of 100 widgets are defective, on average). The next production run will contain 100 widgets. Assume the Poisson distribution is a good fit.

(i) What is the probability that exactly two (2) defects will be produced?
(ii) What is the probability that no more than two (2) defects will be produced (i.e., less than three defects)?
(iii) A defect is considered a “rare event” (i.e., small probability and large number of trials). In this case, the Poisson distribution closely approximates the binomial. Under the binomial distribution, what is the probability of exactly two (2) defects?
(iv) What is the mean and variance of this Poisson distribution?
(v) Can the normal distribution approximate this Poisson distribution? If not, under what circumstances could the normal approximate this Poisson?

Answer:

Here is an EditGrid spreadsheet if you’d like to take a closer look at the Poisson pdf

(i) What is the probability that exactly two (2) defects will be produced?

Given lambda (mean) of 2, under the Poisson distribution P(X=2) = 27.1%.

The numerator of the pdf = (lambda ^ X)exp(-lambda) = (2^2)exp(-2) = 0.541
The denominator of the pdf = 2! = 2
So, the pdf = 0.541/2 = 27.1%

(ii) What is the probability that no more than two (2) defects will be produced (i.e., less than three defects)?

P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 67.7%

This is the nature of a discrete density function: the x-values are independent and discrete. We can add them!

(iii) A defect is considered a “rare event” (i.e., small probability and large number of trials). Therefore, the binomial distribution closely approximates the binomial. Under the binomial distribution, what is the probability of exactly two (2) defects?

=BINOMDIST(2,100,2%,FALSE). The ‘false’ param gives us a pdf instead of a cumulative function.

Note how close this is to the Poisson result!

(iv) What is the mean and variance of this Poisson distribution?

Mean = lambda = 2
Variance = lambda = 2

That’s the funny thing about the Poisson, LAMBDA is BOTH THE MEAN AND THE VARIANCE.

(v) Can the normal distribution approximate this Poisson distribution? If not, under what circumstances could the normal approximate this Poisson?

“No” and “as lambda becomes large”

For a low lambda, like in this example, the Poisson looks like a binomial. As lambda increases (toward infinity), it begins to converge on the normal distribution (and, as they say, the normal approximates the Poisson, or vice )

Early Bird Episode No. 1 - Question 4

2008-01-09T19:13:54-08:00

Question:

Consider a three-asset portfolio consisting of three stocks called A, B, and C. The stocks are weighted, respectively, 20%, 30% and 50% (i.e., weight A = 20%, weight B = 30%, and weight C = 50%). The volatility (standard deviation) of returns for all three assets is 20%. The covariances are as follows: COVAR(A,B) = 0.09 = 9%, COVAR(A,C) = 0.16 = 16%, COVAR(B,C) = 0.01 = 1%. Using matrix math, calculate the three-asset portfolio variance.

Answer:

This is likely a harder question than you would encounter on the exam (not for conceptual difficulty but rather time required to perform the calculation). It is basic matrix application.

Recall that if w is the vector of portfolio weights and Sigma is the covariance matrix, we have a very COMPACT FORMULA for PORTFOLIO VARIANCE:

portfolio variance = (w’)(Sigma)(w)
where w’ is row vector of weights, having been transposed.

In this case, w’ is given by:
[20% 30% 50%]

And the COVARIANCE MATRIX is 3 x 3 and given by:

| 0.04 0.09 0.16 |
| 0.09 0.04 0.01 |
| 0.16 0.01 0.04 |

Note the diagonal of the covariance matrix contains the variances (0.04) which are squared volatilities (0.04 = 20%^2).

The portfolio variance is the product of:
(w’)(Sigma)(w)

Note that matrix multiplication happens to be ‘associative’ so that the order of multiplication does not matter: (AB)C = A(BC).

The worked out answer can viewed in this editgrid spreadsheet.

For reflection:
* The diagonal of the covariance matrix consists of variances. Why? The covariance of a thing with itself is its variance.

* Where is the correlation matrix? It is implicit in the covariance matrix. By the way, what elements are in the diagonal of a correlation matrix? Ones (1). Why? Because the correlation of a thing with itself is 1.0.