We then integrate both sides to obtain e at X t X σ integraldisplay t e as dZ s

We then integrate both sides to obtain e at x t x σ

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We then integrate both sides to obtain e at X t = X 0 + σ integraldisplay t 0 e as dZ s . We then multiply both sides by e - at to obtain the solution X t = X 0 e - at + σ integraldisplay t 0 e - a ( t - s ) dZ s . 3. Define the integral I t = integraldisplay t 0 f ( s ) dZ s . Let 0 = s 0 < s 1 < · · · < s J = t be a grid partitioning [0 , t ] (note that in general s i negationslash = it/J ). We first think of I t as being approximately the left Riemann sum I t J - 1 summationdisplay j =0 f ( s j ) ( Z s j +1 Z s j ) . We shall work out the distribution of this sum and then use this to come up with a conjecture for the distribution of I t . Note that the Riemann sum above is a sum of independent Normal random variables, independence coming from the properties of Brownian motion. Thus, the sum itself is 75
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a Normal random variable. We can thus characterize it’s distribution once we find the mean and variance. First, note that ( Z s j +1 Z s j ) ∼ N (0 , s j +1 s j ). The mean follows from this directly E bracketleftBigg J - 1 summationdisplay j =0 f ( s j ) ( Z s j +1 Z s j ) bracketrightBigg = J - 1 summationdisplay j =0 f ( s j ) E bracketleftbig( Z s j +1 Z s j )bracketrightbig = 0 . To find the variance we note that the variance of a sum of independent random variables is equal to the sum of the variances. Hence, Var bracketleftBigg J - 1 summationdisplay j =0 f ( s j ) ( Z s j +1 Z s j ) bracketrightBigg = J - 1 summationdisplay j =0 f 2 ( s j ) Var bracketleftbig( Z s j +1 Z s j )bracketrightbig = J - 1 summationdisplay j =0 f 2 ( s j ) ( s j +1 s j ) . The above shows us that the Riemann sum we used to approximate I t is normal with mean 0 and the variance given above. As this Riemann sum is only an approximation to the real integral, we have to conjecture something about the true distribution of the integral based on this approximation. We note that our Riemann sum should converge in some sense to I t as J goes to infinity (i.e. as the partition becomes infinitely fine). Looking at the formula for the variance we see that if we take the partition to be as such, the variance will converge to integraldisplay t 0 f 2 ( s ) ds. From this, we find that I t ∼ N parenleftbigg 0 , integraldisplay t 0 f 2 ( s ) ds parenrightbigg . 4. We can now use what we found above for a general integral with respect to Z to find the distribution of X t . We look at the solution of the SDE X t = X 0 e - at + σe - at integraldisplay t 0 e as dZ s and see that it is normally distributed. We thus need to find it’s mean and variance. We do so as follows: E X t = E bracketleftbigg X 0 e - at + σe - at integraldisplay t 0 e as dZ s bracketrightbigg = X 0 e - at Var X t = σ 2 e - 2 at Var bracketleftbiggintegraldisplay t 0 e as dZ s bracketrightbigg = σ 2 e - 2 at integraldisplay t 0 e 2 as ds = σ 2 e - 2 at parenleftbigg e 2 at 1 2 a parenrightbigg = σ 2 2 a ( 1 e - 2 at ) 76
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Therefore, X t ∼ N parenleftbigg X 0 e - at , σ 2 2 a ( 1 e - 2 at ) parenrightbigg 16.2 Trinomial Tree We then construct the trinomial tree approximating X . As usual, let n index time and j index level. We assume as given the parameters a and σ . The time spacing is Δ t , and the level spacing is Δ X , where we choose Δ X = σ t . Let X n,j denote the value of X on the tree at node ( n, j ), with X 0 = 0. At time level n , corresponding to time t
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  • Fall '11
  • COULON
  • Variance, Probability theory, Trigraph, Credit default swap, Wiener process

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