Question 1. Question Solution to final homework!! Friday 5 May Please report any typos immediately for a timely response.

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1 1 Please report any typos immediately for a timely response. Call EW tn W tn 1 near u n 1,...,..., W t1 near u 1 ) LHS for short. Add the quantity W tn 1 W tn 1 into the expected value of the LHS. Apply independent increments property of Weiner process and we get Question 1 LHS = EW tn W tn 1 W tn 1 near u n 1 ) + W tn 1 Finally apply linearity of expectation and the result follows. 2 Recall that for a Weiner process W t Question 2 W t W s Gaussian0, t s) 1) To prove the finite-ness of E W t, it s easy to derive using integration that E W t = 2t π < Next, by independent increments and the fact that W t W s is gaussian with mean 0, EW t W s F s ) = 0. Therefore, EW t F s ) = EW t W s F s ) + EW s F s ) = 0 + W s = W s and so W t is a martingale. Now look at U t : E U t E W t 2 + t = VarW t ) + t = 2t < Also, for s < t EU t F s ) = EW 2 t t F s ) = E[W t W s ) 2 + 2W t W s )W s + W 2 s F s) t = E[W t W s ) 2 ] + 2W s EW t W s F s ) + W 2 s t = VarW t W s ) W 2 s t = t s) + W 2 s t = U s

2 The last two equalities follow from 1). Finally for V t, notice that E V t = exp{2 1 λ 2 t}eexp{λw t }) = 1 < The last equality comes from the fact that the expected value is the mgf of Gaussian0, t). Plug-N -Chug in the exact value of the mgf and they evalute to 1. Finally, for s < t, EV t F s ) = e 2 1 λ 2t Eexp{λW t W s + W s )} F s ) = e 2 1 λ 2t Eexp{λW t W s )} F S )Eexp{λW s } F s ) = e 2 1 λ 2t e 2 1 λ 2 t s) Eexp{λW s } F s ) = e Ws 1 2 λ2s = V s 3 Let T = min{t 0 : W t [a, b]}, p b be the probability of hitting b before a, and similarly for p a. Verify yourself the optional stopping theorem can be used most importantly the uniformly integrability property). Thus, Question 3 Similarly, p a = EW T ) = bp b + a1 p b ) = EW 0 ) = 0 p b = b b a a b a Next, apply optional stopping theorem one more time to see that EU T = EU 0 = 0 EW 2 T ) = ET Note that ET = EW 2 T ) a2 + b 2 < and EW 2 T ) = a2 p a + b 2 p b = ab ET = ab 2

3 4 We assume that gu, t) satisfies gu, h) = πv, s u)gv, t + s)dv 2) Question 4 EZ t+h F t ) = EZ t+h X t = u) = E[gX t+h, t + h) X t = u) = gv, t + h)πv, h u)dv = gu, t) First and last equalities are derived from the fact that X is a Markov process and 2) respectively. For the case gx, t) = x 2 t, πx, s u)gx, t + s)dx = = { } 1 2πs exp x u)2 [x 2 t + s) ] dx 2s { } 1 2πs exp x u)2 2s x 2 dx t + s) = s + u 2 t + s) = u 2 t = gu, t) Similarly for gx, t) = exp { } λx λ2 t 2, use your wonderful integration skills to show that πx, s u)gx, t + s)dx = = gu, t) { 1 exp 2πs } { x u)2 exp 2s λx λ2 t + s) 2 } dx 3

4 5 We can write Question 5 n B = b j B tj for fixed b j = = n b j W tj t j W 1 ) n n b j W tj W 1 b j t j Sum and difference) of Gaussians is Gaussian. So B is Gaussian. This means that any linear combination finite) of B t is gaussian. So the collection B t1,..., B tn ) is multivariate Guassian and B t is a Gaussian process. The mean function is EB t ) = EW t ) tew 1 ) = 0 and the autocovariance function is given by CovB t, B s ) = E[W t tw 1 )W s sw 1 )] = EW t W s ) + stew 2 1 ) tew 1 W s ) sew 1 W t ) = s + st ts st = s1 t) 6 For simplicity, let Question 6 B k = W k2 n t W k 1)2 n t Since W t is a standard Weiner process, B k iid Gaussian0, 2 n t). Then by the Law of Large Numbers, B k 2 n E B k ) ) = n+1) t 1/2 as n π So it is proven that B k 4

5 Now let nk and U nk be as defined in the assignment. Equation 7) in the assignment can be rewritten as lim n U nk + t2 n ) = t or equivalently U nk 0 The following are true: EU nk = E 2 nk t2 n ) = 0 recall nk iid Gaussian0, 2 n t) EU 2 nk ) = E 4 nk + 2t 2 n 2 nk + t2 2 2n ) = 3t 2 2 2n 2t 2 2 2n + t 2 2 2n = 2t2 2 2n 0 as n and so L 2 convergence follows. The last equality follows from the hint. By Chebyshev s inequality, we have that 2 n ) P U nk > ɛ [ 2 ) ] n 2 E U nk ɛ 2 n 2 n ) P U nk > ɛ = 2t2 ɛ2 n 2t2 ɛ 2 due to iid property < Finally, apply the Borel-Cantelli Lemma, 2 n ) P U nk > ɛ i.o. = 0 and so for large n, 2 n ) 2 n ) P U nk < ɛ = 1 P lim U n nk < ɛ = 1 Then, we have that P lim n i=1 U nk = 0 ) = 1 implies U nk a.s. 0 5

6 7 Question 7 ) ) P max W s c = 1 P max W s < c 0 s T 0 s T ) = 1 P max W s c 0 s T = 1 P W T failed to pass c and W T < c) = 1 [P W T < c) P W T < c and W T passed c)] = 1 [P W T < c) P W T > c and W 0 = 0)] ) ) [ )] c c c = 1 Φ + 1 Φ = 2 1 Φ T T T Third to last equality follows from reflection principle. 8 For first case, let W be the waiting time so that 0 W. It can be easily shown using basic definition of expected value for continuous random variables that Question 8 EW = 2 For the second case, the mean measure is Leb and for waiting time W PW > t) = PNA) = 0 where A = 0, t)) = exp LebA) ) = exp t ) This gives the kernel of Exponential 1 ) and so EW =. 6

7 9 Fix some arbitrary point and let D be the distance from the point to the closest tree. Further let A r be the circle with radius r and centered at the fixed point. Then Question 9 λleba r ) = πr 2 λ NA r ) Poissonπr 2 λ) Then we have that PD > r) = PNA r ) = 0) = exp πr 2 λ) One-minus it and we get the CDF of D. For the second part, it s typical to treat the line of sight as a rectangle with arbitrary length l and width 2a. If there is a tree in this rectangular area, call it A, the view to the east is blocked. So we have λleba) = 2alλ NA) Poisson2arλ) Thus, PD > l) = PNA) = 0) = exp 2alλ) D Exponential2aλ) and so ED = 2aλ) 1 10 P NA 1 ) = a 1... NA k ) = a k ) k i 1 = P NA 1 ) = a 1 ) P NA i ) = a i {NA j ) = a j } i=2 Question 10 k = P NA 1 ) = a 1 ) P NA i ) NA i 1 ) = a i a i 1 ) i=2 k = P NA 1 + t) = a 1 ) P NA i + t) NA i 1 + t) = a i a i 1 ) i=2 k i 1 = P NA 1 + t) = a 1 ) P NA i + t) = a i {NA j + t) = a j } i=2 = P NA 1 + t) = a 1... NA k + t) = a k ) Second equality and third equality follow from independent increments and stationary increments respectively. 7

8 11 Given regions A and B, this measure characterizes the process on overlapping parts of A and B. In the stationary case, it is 0. It is also 0 for Poisson process. To see this, we know that Question 11 CovNA), NB)) = 2 1 [VarNA)+NB)) VarNA)) VarNB))] We know NA) and NB) are Poisson with rates λleba) and λlebb) respectively. Now use the mgf of NA) and NB) to derive the mgf of NA) + NB). It turns out it is the mgf of a PoissonλLebA) + λlebb)) Thus, CovNA), NB)) = 2 1 [VarNA)+NB)) VarNA)) VarNB))] = 0 12 Suppose the Poisson random measure has mean measure λleb ) and hence the avoidance measure is given by Question 12 V A) = PNA) = 0) = e λleba) 13 Define A t = {points p : p t}. Denote the overall intensity λ. By the definition of homogeneous Poisson process, NA t ) is Poisson with rate λv At where V At is the volume of A t. But Question 13 V At = v d t d Thus, Kt) = E[NA t)] λ = λ v d t d λ = v d t d 8