Answer Happy

3. Let (Xt) te[0,1] be the solution to the one-dimensional SDE dX; = a(X4)dt + b(X4)DW, for t€ (0,T) and Xo = 3o E R where (W.)e>, is the standard Brownian motion and a,b are one-dimensional deterministic functions. Take := {ti = ih : h:=T/N, i=0,...,N} as the uniform time grid of (0,T) with fixed stepsize h for some fixed natural number N >1 and let {X}} be the standard Euler scheme approximation of X over 7. Namely, set X = Xo = to and for k = 0,...,N-1 we define xk+1 = x + a(x)h +6(X)AWk, where AW:=W+x+1 - W We now introduce an alternative scheme called the Weak Euler Scheme. Set Yo = Xo = Xo and for n = 0,...,N - 1 we define Yn+1 = Yn+anh + bnr Vh, with an := a(Y) and bn := b(Y), where &r are i.i.d. Binomially distributed random variables such that P[En = +1] = { for any n. Important note: Recall that the filtration generated by the Brownian motion is (F1).>0. Critically, for any k € {0, ... ,N}: the ak and bk are Fl-measurable random variables; and Ex is independent of Fix (and AWx). (a) Assume the function a is uniformly bounded, i.e., for some M > 0 for all x E R we have that |a(2) < M. Consider the following definition of Weak consistency. Definition (Weak consistency). A numerical scheme is called Weakly Consistent if it satisfies the following two conditions (W1) him E [Yn+1- E h an = 0 0 */7.) --))- elle [wat 5.1 - 6) ] 1= (W2) lim E E (Yn+1 - Y,)? h F-62 = 0 h0 (i) Start by calculating E[En] and E[8] and show that the Weak Euler Scheme is weakly consistent (12 marks] (ii) Is it possible to generalise the choise of En and still have a weakly consistent scheme? Justify your answer. [7 marks] (b) A scheme is called Strongly Consist if it satisfies Condition W1 of Weak consistency and also lim E h→0 [ |-:1 – V. – E Vnt. –V| F.) – b,aw.r] = = 0. Is the Weak Euler Scheme strongly consistent? To justify your answer you may assume the function b(x) to be constant equal to be R.