Monte Carlo Methods. Matt Davison May University of Verona Italy

Transcription

1 Monte Carlo Methods Matt Davison May University of Verona Italy

2 Big question 1 How can I convince myself that Delta Hedging a Geometric Brownian Motion stock really works with no transaction costs? What are the impact of transaction costs? What is the impact of changing the model from GBM?

3 Big insights What s a good way to do quadrature in high dimensions? How do stochastic differential equations really work?

4 All of these questions Can be contributed to by the study of simulation methods. We will begin our story with a very simple problem, however.

5 Simple case study You are a life insurance provider You have written (very simple) term life insurance policies on 1000 very similar people (75 year old non smoking men, say) The policy pays $100,000 if the policy holder dies in the next year, otherwise nothing. Your actuaries tell you that the probability of this happening, for a given insured man, is 5%.

6 What insurance do you charge? So the insurance you charge is: 5.5% x $100,000 = $5500 5% covers expected loss, 0.5% is partly a profit and partly to cover risk. What risk do you have?

7 Let s simulate Let s simulate what happens. There are statistical short cuts to make this simpler, but let s just go really naïve: For each of 1000 men, we generate a random variable that with probability p = 5% is 1, and with probability 1 p = 95% is 0. (a Bernoulli random variable). It is easy to do this even in Excel, but easier to If we do this, once, we get 39 deaths

8 That is good, right? We sold 1000 policies for $5500 each That s $5.5 million We had to pay out 39 times, $100,000 each time That s $3.9MM Profit is $1.6MM What a great business! Where do I sign up?

9 But that was just one run That was just one run. We know that on average we should have 50 deaths. This time we had fewer than 50 which is good, but perhaps we might have more than 50? So let s run the simulation 100 times. We get:

10 Matlab code function Y = failure(n,m,p) % Returns an N x m matrix of iid Bernoulli(p) draws X = rand(n,m); Y = zeros(n,m); for i = 1: N for j = 1:m if X(i,j) < p Y(i,j) = 1; end end end

11 Sampled 100 times

12 Summary Stats This is 100 realizations of claims on next year s insurance book The maximum number of claims was 67 (which would have us lose $1.2 MM why?) The minimum number of claims was 32 The mean number of claims was 48.3 And the median number of claims was 48 8/100 realizations had more than 60 claims

13 What to make of this 17 of the 100 realizations had > 55 claims (so lost money 8 of 100 had > 60 claims 3 of 100 had > 65 claims (so lost > $1MM).

14 How to reduce risk? What if we sold 10,000 $10,000 policies instead of 1000 $100,000 policies? So run our failure(n,m,p) code again but with N = 10,000 (m = 100, p = 0.05) Then Y = failure(10000,100,0.05) z = sum(y) (so z is a 100 element vector) Then max(z),min(z), mean(z), median(z) etc for summary stats, hist(z) for histogram

15 New histogram

16 Some conclusions Here we are making money if we have < 550 claims Sales = x 10,000 policies x $10,000 per policy = $5.5 million (same as last time) Payouts = claims x $10,000 The max number of claims in 100 runs was 555, for a loss of $50,000, not too bad.

17 Business discussion If you change the size of the policies like this can you still sell the same number? Will it be to the same type of people? If you want to scale the business up, (i.e, sell 10,000 $100,000 policies) can you afford to? How much capital do you need? How can you protect against your loss?

18 Reinsurance Maybe stick with 1000, $100,000 policies but protect against risk by buying a reinsurance contract that covers all claims above some threshold. (NB: in real life often only say 90%, why?) So the reinsurance policy pays 0.9* $100,000 *Max(z-thresh,0) Where z is the number of claims made.

19 Some code function [q,l] = insurance(l,n,m,p,thresh) % Constructs two L vectors. q gives, for each trial, the number of exceedances of % thresh claims in m trials of a book of N policies, prob (claim) = p. % l gives the total number of claims exceeding thresh for each trial for i = 1:L Y= failure(n,m,p); z = sum(y); q(i) = 0; l(i) = 0; for k = 1:m if z(k) > thresh q(i) = q(i) + 1; l(i) = l(i) + z(k) - thresh; end end end

20 Running this [q,l] = insurance(5,1000,100,0.05,60) q = l = So this means that in the first run, 3 of 100 years had exceedances of 60 claims, and the total exceeded sum over 60 claims was 27 The average of l over a big enough run should price the reinsurance contract.

21 Reinsurance >> [q,l] = insurance(500,1000,100,0.05,60); >> mean(l) ans = This means that the expected number of claims over 60 in 100 trials is about 26. Note that mean(q) = 6.772, (of 100 trials) so the reinsurance contract only pays off about 6.8% of the time, and when it does it covers claims above 60. Each claim is worth $100,000. I estimate that the reinsurance contact should be worth 0.05% of the total book of business (100 million). So $50,000.

22 What if we increase the strike >> [q,l] = insurance(500,1000,100,0.05,65); >> mean(l) = (note max(l) = 26, min(l) = 0) So total payout is x 100,000 = 4500 per year. >> mean(q) =1.4620, max(q) = 5, min(q) = 0). Reinsurance only pays out about 1.5% of the time >> mean(l)/mean(q) = (This contract is much cheaper than before) When it does pay out it pays for about 3.14 claims.

23 Test If thresh = 30, we should the reinsurance contract is mostly like the insurance contract except it is paying only for after 30. >> [q,l] = insurance(5,1000,100,0.05,30) q = l =

24 Purpose of Simulation For calculation But also for idea generation To reinforce the idea that the average behaviour is not the only thing that matters Have to watch max and min as well.

25 We will discuss: In Matlab there is a rand() function that generates U(0,1) variables. From that we can generate other variables as required. How does rand work? When have you done enough simulations? We begin with the when is enough question as it is the most important.

26 How much work do you need to do to find an unfair die? Let s simulate rolling a regular 6 sided die We are equally likely to roll 1,2,3,4,5 or 6. What is the average roll? Of course we can work this out by hand and pretty quickly arrive at 3.5, but let s see what the simulations tell us.

27 Results Group Avg (10 sim rolls) Avg (100 sim rolls) Avg 1000 sim rolls) Average STD

28 Take home lessons The simulation isn t the same as the analytic answer even when we do 10 groups of 1000 simulations (10,000 simulations) But it is getting pretty close (It gets better with N here, but not always). The Std Dev measure of how far the average is from where it should be seems to be scaling slower than linear.

29 Theory Suppose X 1, X 2, X N are identically distributed random variables with finite mean µ and finite variance V independent. Write X = (1/N)(X 1 + X X N ) E(X) = (1/N)(µ + µ + + µ) = (1/N)(Nµ) = µ Var(X) = (1/N 2 )(V + V + V) = (1/N 2 )(NV) = V/N This suggests that the standard deviation of the estimator scales as (1/ N)

30 Even simple simulations have uses What if a die is loaded to be much more likely to roll a six (25% likely; 15% likely to turn up on the other five faces) Mean is then 0.25* *( ) = *15 = = 3.75 If we simulate rolling that die as before we get the following outcomes: check to see if you could convince yourself they were different from 3.5, the mean of a fair die.

31 Stats for biased die Group Avg (10s) Avg (100s) Avg (1000s) Average STD

32 Conclusions? After 10 or even 100 rolls it will be very hard to conclude it has been tampered with even it if has been with 1000 rolls it starts to be evident with 10,000 a near certainty. This sort of quick and dirty calculation is very good for determining the answer to tricky statistics problems And for convincing your managers that they are correct.

33 How do we do it? We need to find a way to simulate IID random numbers The typical route is by first simulating random variables that are equally likely to return any number between 0 and 1 These are so called U(0,1) random variables Then we study how to use transformations of (sets of) U(0,1) r.v.s to simulate any type of r.v we want.

34 Simulating U(0,1) random variables What I m about to share with you seems strange at first But it does appear to (empirically) work. The idea is that we will create a stream of random numbers using a deterministic computer. In fact, this will be a big advantage for running our experiments. First let s do some theory

35 A detour into Chaos theory In the 1980s a subject called Chaos theory was all the rage for about a decade. One of the ideas of chaos theory was that simple deterministic maps could create complicated, and random looking, dynamics. The (often unjustified) corollary what that complicated and random looking behaviours had simple causes that could be discovered. This rarely turned out to be the case. But lots of nice math was discovered, or rediscovered.

36 Discrete Maps One sub-field of chaos theory was Discrete Maps These were maps X n+1 = S(X n ) of the real line, or some subset (say [0,1]) of the real line into itself. These maps would be iterated and the result would be a sequence with full deterministic, but very interesting, properties.

37 Shift maps X n+1 = (rx n )mod 1 This maps [0,1) into [0,1). In fact a fixed point of the map is 0, so let s just consider mapping (0,1) into (0,1) The next slide depicts this for r = 4. The mapping is many to one Binary shift map: r = 2 Decimal shift map: r = 10

38 Shift Map (4x)mod

39 Shift maps and SDIC Shift maps display sensitive dependence on initial conditions Under the r = 10 decimal shift map, the initial points X0 = and Y0 = Get shifted to and and and 0.779, 0.70 and 0.79, then 0 and 9.

40 Mapping densities under discrete maps Let s return to the general setup for a minute X n+1 = S(X n ), is a map from X to X. If X is a subset of R n, then we can associate to the (nonlinear) map F a (linear) map called the Perron-Frobenius operators. (This is a functional map so is infinite dimensional, so what we gain in linearity we lose in dimensionality).

41 Density Maps (II) Now let A be a subset of X. Then A f n+1 (s)ds (= A Pf n (s)ds =) S-1(A) f n (s)ds This says that all the points mapped from time n to to time n+1, if they are in A at time n+1, must have been in the pre-image of A under S at time n.

42 Applied to the shift map So, if we are going to apply A Pf n (s)ds = S-1(A) f n (s)ds with X = (0,1), S(x) = (rx)mod1, and A = (0,x) then S -1 (0,x)= (0,x/r) U (1/r, (1+x)/r) U (2/r, (2+x)/r) U U( (r-1)/r, (r+x-1)/r) So 0x Pf n (s)ds = 0 x/r f n (s)ds + 1/r x+1/r f n (s)ds + + (r-1)/r (r+x -1)/r f n (s)ds

43 Solving the equation Differentiating both sides with respect to x yields the functional equation f n+1 (x)= Pf n (x) = 1/r [ f n (x/r) + f n ((x+1)/r) + f n ((r+x-1)/r)] (**) Note that f*(x) = on [0,1], 0 otherwise is a fixed point of this equation; the averaging inherent in (**) will drive initial functions to the uniform distribution.

44 Shift maps for generating random numbers. Motivated by this there has been much development of Linear Congruential Generators, or LCGs. Will just scratch the surface here. These are maps on positive integers that are very reminiscent of the shift maps we just discussed.

45 Largest class of LCGs x n = (ax n-1 + c) mod m Initialize the generator with a seed x 0. Example: x n = (5x n-1 + 3) mod 8, x 0 = 3. Then we get 2,5,4,7,6,1,0,3 Once it reachs x 8 = 3, the pattern simply repeats again.

46 Some features Period of the generator can be at most m-1 (there are only m-1 distinct positive natural numbers when we are working mod m), So, in practice, m should be very large. Of course the period of the generator could still be less than m. For so-called multiplicative generators (in which c = 0); x n = (ax n-1 ) mod m, we can say that:

47 Max Period theorem If m is prime, the multiplicative congruential generator x n = (ax n-1 ) mod m has maximal period m -1 iff (a i ) mod m 1 for all i = 1..m bit machines often use the Mersenne prime m = For this value of m many values of a, including 7,16807,39373, and many others Produce full period generators.

48 Getting from naturals to [0,1] Take numbers from {1,2, m-1}, add ½, divide by m. This guarantees numbers between 0 and 1 without actually generating 0 and 1. Much more work on the details of this; for example Pierre L Ecuyer in Montreal Canada

49 IID? These are however definitely not independent quite the opposite. But they look independent can fool many tests. (Also fail many sophisticated tests) The fact that they are dependent means you can generate the same sequence of random numbers over and over again by using the same seed ; this can be very useful.

50 Transforming U(0,1) numbers Most random numbers arising in financial applications are not U(0,1). Earlier today we talked about discrete distributions (fair or biased die rolls). Could also have exponential distributions, normal distributions, and many more. How do we transform U(0,1) numbers into numbers distributed another way?

51 Generating discrete RVs A big class of discrete RVs are the fair n sided die models. Of course our simulated dies need not be related to the platonic solids! These are easy to simulate from Uniforms: Rand(6) = ceiling(6*u(0,1),1) etc. This is how the fair dice rolls were done at the beginning of class.

52 Using inverse CDFs If F is an arbitrary cumulative distribution function and U is Uniform(0,1) then X = F -1 (U) has cumulative distribution function F(x). For instance the plot on the following page for the exponential distribution with parameter θ

53 Θ = 2

54 Generating exponentials The exponential distribution with parameter θ Has pdf f(x) = (1/θ)exp(-x/θ) x 0, = 0, x < 0. Thus F(x) = 1 exp(-x/θ), x 0. So F -1 (U) = -θln(1-u) This is equivalent to θln(u), since U and 1-U have the same distribution.

55 This is very easy to implement; even in Excel With θ = 2 and 100 points we already get a very nice Quantile-Quantile (QQ) plot of empirical vs. theoretical distribution See next slide

56 QQ plot for simulated exponentials

57 Simulating Normals As usual with normals the trick is to move into two dimensions (think of trick for solving showing the integral of the normal pdf is 1). If X and Y are jointly normal with correlation zero, then we can write r 2 = x 2 + y 2 tanθ = y/x Then R has pdf r exp(-r 2 /2) for r > 0 And Θ has pdf 1/2π for 0 < θ < 2π

58 Box-Muller Normal R 2 has exponential distribution with mean of 2 Θ is uniform on (0,2π) Then (X,Y) = ( R cos Θ, R sin Θ) are jointly exponential. R = (-2ln(U)) Θ = 2πV Where U and V are U(0,1) variables.

59 QQ plot (N=100, N(0,1)

60 Comments Not quite as good as the Exponential outcome. But the 100 points we used isn t very many.

61 Accept-Reject methods Working with inverse CDFs isn t always practical it may not be analytically tractable. Sometimes we use accept-reject. For example, what if we want to generate (x,y) points uniformly sampled from the region of the unit circle? First simulate (x,y) as two U(0,1) draws.

62 Reject step Then check to see if the point so generated lies inside the unit circle: If x 2 + y 2 1 keep the point, otherwise reject it. The points that remain are still uniformly distributed on the disc. This method is wasteful of points, though, keeping only π/4 of those generated. Also a (very inefficient) way of estimating π!

63 Accept-Reject tricks You want to draw the initial points from a region as close as possible to the final one. Then most points are used.

64 Bernoulli Sampling Take a minute to consider a very simple simulation. You have N draws of a Bernoulli(p) random variable, i.e. a random variable with value 1 a fraction p of the time, otherwise a value 0. Here N could be the number of loans granted, each of value X, and p the annual probability of default.

65

66 Monte Carlo Integration Suppose we want to solve 0 1 f(x)dx Sample x as uniform (0,1); x i = 1 N Then estimate with 1/N k=1 N f(x k ) This will converge at the rate 1/ N

67 On the other hand, if we use a trapezoid rule with N panels we ll get 0 1 f(x)dx = (1/2N) {[f(0) + f(1/n)] + [f(1/n) + f(2/n)] + + [f((n-1)/n) + f(1)] = (1/2N){f(0) + f(1)} + (1/N){f(1/N) + f(2/n) + + f ((N-1)/N)} The error in each panel here is order ½ h 2 f (z i ), where z * i is some point between (i-1)/n and i/n, and where h = 1/N. Add that up we have ½ N (1/N)2 *the average curvature of the function over [0,1]

68 MC wins in lots of dimensions The average curve is 1/N * biggest curvature value on each interval. So error = ½ avecurve *(1/N). This is faster than Monte Carlo, and we don t need to do any tech for it. In 2D, however, to get the same grid point coverage you need N 2 grid points; N for each direction. So the order of safety appears the same as Monte Carlo. Once we get to 3D we get third root convergence, which is slower for traditional quadrature than for MC..

69 Why? High dimensional spaces are big The Monte Carlo goes ahead and checks them all over the place at least.

70 Exercises Reproduce some stuff Use simulation to see how many trials one would need to do before consider a coin to be biased. The particular coin being considered here has p(heads) = 55%, p(tails) = 45%.