American Binomial Model in q

American-style options may be exercised prior to expiry. To price them on a binomial tree, the full tree has to be constructed. The prices are then computed at each node through a backward reduction process starting from the prices at maturity. At each node, we take the maximum of the payoff at that node versus the price implied by the binomial model.

Constructing a binomial price tree is relatively easy in q:

```BPTree:{[n;S;u] n{(x*y 0),y%x}[u]\1#S}  / binomial price tree
```

where n is the depth of the tree, S is the current price and u is the scale of the up-move. We simply let the scale of the down-move be 1/u and hence the y%x in the expression.

The general binomial model can then be implemented as follow:

```GBM:{[R;P;S;T;r;b;v;n]                  / General Binomial Model (CRR)
t:T%n;                                 / time interval
u:exp v*sqrt t;                        / up; down is 1/u
p:(exp[b*t]-1%u)%(u-1%u);              / probability of up
ptree:reverse BPTree[n;S;u];           / reverse binomial price tree
first R[exp[neg r*t];p] over P ptree }
```

where R is a reduction function, P is the payoff function, S is the current price, T is the time to maturity, r is the risk-free rate, b is the cost of carry, v is the volatility and n is the depth of the tree. For American and European options, the reduction functions may be expressed as:

```American:{[D;p;a;b] max(b;D*(-1_a*p)+1_a*1-p)}
European:{[D;p;a;b] D*(-1_a*p)+1_a*1-p}
```

where D is the discount factor and p is the probability of an up-move. Consequently, we can express the American and European binomial models simply as:

```ABM:GBM[American]
EBM:GBM[European]
```

Testing the code on an American vanilla put option (strike = 102; price = 100; time to maturity = 1 year; risk-free rate = cost of carry = 8%; volatility = 20%; depth of tree = 1000):

```q) VP:{[S;K]max(K-S;0)}    / vanilla put: max(K-S,0)
q) \t show ABM[VP[;102];100;1;0.08;0.08;0.2;1000]
6.2215
31
```

It took 31 ms to compute. Pretty nice for so little code.

Note: It turns out that this implementation of EBM is faster than the one in the previous post. The reason is that I avoided using the expensive xexp function this time round. Otherwise, the previous implementation should be faster since it only computes the payoffs at maturity and not the intermediate nodes.

European Binomial Model in q and Python

In the previous post, we created a binomial probability mass function (pmf). We can use that to easily evaluate European-style options:

```EBM:{[P;S;K;T;r;b;v;n]                 / European Binomial Model (CRR)
t:T%n;                                / time interval
u:exp v*sqrt t;                       / up
d:1%u;                                / down
p:(exp[b*t]-d)%(u-d);                 / probability of up
ns:til n+1;                           / 0, 1, 2, ..., n
us:u xexp ns;                         / u**0, u**1, ...
ds:d xexp ns;                         / d**0, d**1, ...
Ss:S*ds*reverse us;                   / prices at tree leaves
ps:pmf[n;p];                          / probabilities at tree leaves
exp[neg r*T]*sum P[Ss;K]*ps }
```

Note that P is the payoff, S is the current price, K is the strike price, T is the time to maturity, r is the risk-free rate, v is the volatility, b is the cost of carry and n is the depth of the binomial tree. The Python version using NumPy and SciPy actually looks quite similar:

```def EuropeanBinomialModel(P, S, K, T, r, b, v, n):
n = int(n)
t = float(T)/n                              # time interval
u = np.exp(v * np.sqrt(t))                  # up
d = 1/u                                     # down
p = (np.exp(b*t)-d)/(u-d)                   # probability of up
ns = np.arange(0, n+1, 1)                   # 0, 1, 2, ..., n
us = u**ns                                  # u**0, u**1, ...
ds = d**ns                                  # d**0, d**1, ...
Ss = S*us*ds[::-1]                          # prices at leaves
ps = binom_pmf(ns, n, p)                    # probabilities at leaves
return np.exp(-r*T) * np.sum(P(Ss,K) * ps)
```

As we can see, both code has no explicit loops. This is possible in Python as NumPy and SciPy are array-oriented. NumPy and SciPy’s idea of “broadcasting” has some similarity with k/q’s concept of “atomic functions” (definition: a function f of any number of arguments is atomic if f is identical to f’).

Binomial distribution in q

Recently I was using SciPy’s `scipy.stats.binom.pmf(x,n,p)`. I though it would be great if I could have such a function in q. So a simple idea is to construct a binomial tree with probabilities attached. Recalling that a Pascal triangle is generated using `n{0+':x,0}\1`, I modified it to get:

```q)pmf:{[n;p]n{(0,y*1-x)+x*y,0}[p]/1#1f}
q)pmf[6;0.3]
0.000729 0.010206 0.059535 0.18522 0.324135 0.302526 0.117649
q)sum pmf[1000;0.3]
1f
```

What is great about this method is that it is stable. Compared to SciPy 0.7.0, it was more accurate too (it is a known issue that older SciPy has buggy binom.pmf):

```>>> scipy.stats.binom.pmf(range(0,41),40,0.3)[-5:]
>>> array([3.33066907e-15, 0.00000000e+00, 0.00000000e+00, 0.00000000e+00, 1.11022302e-16])
q)-5#pmf[40;0.7]
3.293487e-15 1.52594e-16 5.162955e-18 1.134715e-19 1.215767e-21
```

Unfortunately this method is too slow for large n. For large n, we need more sophisticated methods. For the interested reader, take a look at Catherine Loader’s Fast and Accurate Computation of Binomial Probabilities paper and an implementation of a binomial distribution in Boost

Posted in coding, k&q, python. Tags: , , . 1 Comment »

Gray code in q

It is easy to construct binary n-bit Gray code in q using the recursive reflection-prefixing technique:

```q)gc:{\$[x;(0b,/:a),1b,/:reverse a:.z.s x-1;1#()]}
q)show gc 4
0000b 0001b 0011b 0010b 0110b 0111b 0101b 0100b
1100b 1101b 1111b 1110b 1010b 1011b 1001b 1000b```

It is also possible to construct the above iteratively using the formula $n \oplus \lfloor n/2 \rfloor$:

```q).q.xor:{not x=y}
q)gc_iter:{(0b vs x) xor (0b vs x div 2)}
q)show (-4#gc_iter@) each til 16
0000b 0001b 0011b 0010b 0110b 0111b 0101b 0100b
1100b 1101b 1111b 1110b 1010b 1011b 1001b 1000b```

To check that indeed exactly one bit is flipped each time:

```q)check:{x[0] (sum@xor)': 1_x}
q)check gc 5
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1```

To identify the position of the bit that was flipped:

```q)pos:{raze x[0] (where@xor)': 1_x}
q)pos gc 5
4 3 4 2 4 3 4 1 4 3 4 2 4 3 4 0 4 3 4 2 4 3 4 1 4 3 4 2 4 3 4```

If we think about it, there is no reason why we have to prefix. We could do suffix as well:

```q)gc:{\$[x;(a,\:0b),(reverse a:.z.s x-1),\:1b;1#()]}
q)show gc 4
0000b 1000b 1100b 0100b 0110b 1110b 1010b 0010b
0011b 1011b 1111b 0111b 0101b 1101b 1001b 0001b
q)check gc 5
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
q)pos gc 5
0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 4 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0```

In fact, if we are only interested in the positions that need to be flipped, we can use this instead:

```q)gcpos:{\$[x;a,n,a:.z.s n:x-1;()]}
q)gcpos 5
0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 4 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0```

Such a sequence of positions is useful if we are using Gray code to efficiently enumerate the non-zero points spanned by a set of basis vectors:

```q)basis:(1100000b;0111001b;0000011b)
q){x xor y} scan basis gcpos count basis
1100000b
1011001b
0111001b
0111010b
1011010b
1100011b
0000011b```

Update (20090927): Once again, Attila has beaten me at q-golf :-) Here is his formulation:

`gc:{x{(0b,/:x),1b,/:reverse x}/1#()}`