STOCHASTIC METHODS IN FINANCE 2021–22 STAT0013
Suggested Solutions to Exercises 5
(a) From the properties of BM, the mean at time T is zero, and the variance is T.
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(b) If the process is fixed at k at t = 1, we need to consider the increment between time t = 1 and time t = 3. From the properties of BM, the distribution of the increment W3 − W1 will be normal, with mean zero and variance equal to the length of the time period, which is 3−1 = 2. Thus, conditional on W1 = k, the mean of W3 is k + 0 = k, and the variance is 2.
We have that
S6 = (S6 −S3)+(S3 −S0)+S0
By the properties of the generalised BM we know that
(S3 −S0)∼N(2×3,32 ×3) (S6 −S3)∼N(3×3,42 ×3)
Also, the two above increments are independent. It follows that: S6 ∼N(2×3+3×3+5,32 ×3+42 ×3)
C1 ∼N(C0 +4×0.5,4×4)
P[C1 < 0] = P C1−C0−2 < −C0−2 = N(−C0−2)
S6 ∼N(20,75) of C1 (i..e., the cash position after 1 year) is
We conclude that
Let the company’s initial cash position be C0. Then the distribution
It follows that
where N(·) is the cdf of the standard normal distribution. From the table of the standard normal distributionm we have
N(−C0−2) = 0.05 ⇔ −C0−2 = −1.6499 44
Solving for C0 we obtain C0 = 4.5996. Thus, the initial cash position must then approximately £4.6 millions.
4. We need to verify for Zt the properties that define BM. Since Wt is BM, it has continuous paths by definition, and clearly Zt = −Wt also has continuous paths. We also have that Z0 = −W0 = 0, thus {Zt} satisfies another requirement of BM.
Consider the increments, for 0 < s < t,
Zt −Zs =−Wt +Ws =−(Wt −Ws)
We know that Wt −Ws ∼ N(0,t−s), so also Zt −Zs ∼ N(0,t−s). Thus, {Zt} satisfies the BM property of normal increments.
Since {Wt} has independent increments, we know that Wt2 − Wt1 is independent of Ws2 − Ws1 for s1 < s2 ≤ t1 < t2. This means that −Zt2 + Zt1 is independent of −Zs2 + Zs1 , thus Zt2 − Zt1 is independent of Zs2 − Zs1 . In other words, {Zt} has independent increments.
Therefore {Zt} is a BM, as it possesses all the properties of a BM.
5. The pdf of Bt is equal to
1 x2 f(x;t)=√2πtexp −2t .
Taking partial derivatives with respect to t and x, we obtain
∂f = x2 −tf, ∂f =−xf, ∂2f = x2 −tf =2∂f,
∂t 2t2 ∂x t ∂x2 t2 ∂t so that the given diffusion equation is satisfied.
6. Without loss of generality we assume that s < t so that min{s, t} = s.
(a) Long Approach - Not Recommended: Using the properties of BM and conditioning on the value of the process at time s, we can write the joint pdf of Wt, Ws as
fWs,Wt (x, y) = p(s, 0, x) p(t − s, x, y)
where p(·, ·, ·) is the pdf of a normal distribution N (μ, t), that is
as required.
E[Ws Wt ] = min{s,t}
2t p(t, μ, x) = √1 e− (x−μ)2
Thus, we have
E [ Ws Wt ] =
xy p(s, 0, x)p(t − s, x, y)dxdy ∞ ∞
∞∞ −∞ −∞
xp(s,0,x) yp(t−s,x,y)dy dx −∞
The inner integral can be written as
yp(t−s,x,y)dy = (x+u)p(t−s,x,x+u)du
where we have used the substitution u = y − x in the first step, and properties of the normal density thereafter. Thus, returning to the expression for E [ Ws Wt ], we have
where we have again used the properties of the pdf of the normal distribution, and have corresponded the integrals to simple mo- ments of an appropriately parameterized normal distribution. As we assumed that min{s, t} = s, we have indeed obtained
= (x+u)p(t−s,0,u)du
= x p(t − s, 0, u)du + u p(t − s, 0, u)du −∞ −∞
E[Ws Wt ] =
x2p(s,0,x)dx = s
(b) Fast Approach - Highly Recommended: We use iterative conditional expectation, and obtain that (again, we have assumed that s < t)
E[Ws Wt ] = EE[Ws Wt |Ws] = EWs E[Wt |Ws ]
= E [ ] = s
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