Consider a symmetric simple random walk ( Z n ) n ∈ Z + \left(Z_{n}\right)_{n \in \mathbb{Z}^{+}} ( Z n ) n ∈ Z + taking values in statespace I = h Z 2 ≡ { ( i h , j h ) : i , j ∈ Z } I=h \mathbb{Z}^{2} \equiv\{(i h, j h): i, j \in \mathbb{Z}\} I = h Z 2 ≡ { ( i h , j h ) : i , j ∈ Z } , where h ≡ N − 1 ( N h \equiv N^{-1}(N h ≡ N − 1 ( N an integer ) ) ) . Writing Z n ≡ ( X n , Y n ) Z_{n} \equiv\left(X_{n}, Y_{n}\right) Z n ≡ ( X n , Y n ) , the transition probabilities are given by
P ( Δ Z n = ( h , 0 ) ) = P ( Δ Z n = ( 0 , h ) ) = P ( Δ Z n = ( − h , 0 ) ) = P ( Δ Z n = ( 0 , − h ) ) = 1 4 , P\left(\Delta Z_{n}=(h, 0)\right)=P\left(\Delta Z_{n}=(0, h)\right)=P\left(\Delta Z_{n}=(-h, 0)\right)=P\left(\Delta Z_{n}=(0,-h)\right)=\frac{1}{4}, P ( Δ Z n = ( h , 0 ) ) = P ( Δ Z n = ( 0 , h ) ) = P ( Δ Z n = ( − h , 0 ) ) = P ( Δ Z n = ( 0 , − h ) ) = 4 1 ,
where Δ Z n ≡ Z n − Z n − 1 \Delta Z_{n} \equiv Z_{n}-Z_{n-1} Δ Z n ≡ Z n − Z n − 1 .
What does it mean to say that ( M n , F n ) n ∈ Z + \left(M_{n}, \mathcal{F}_{n}\right)_{n \in \mathbb{Z}^{+}} ( M n , F n ) n ∈ Z + is a martingale? Find a condition on θ \theta θ and λ \lambda λ such that
M n = exp ( θ X n − λ Y n ) M_{n}=\exp \left(\theta X_{n}-\lambda Y_{n}\right) M n = exp ( θ X n − λ Y n )
is a martingale. If θ = i α \theta=i \alpha θ = i α for some real α \alpha α , show that M M M is a martingale if
e − λ h = 2 − cos ( α h ) − ( 2 − cos ( α h ) ) 2 − 1 e^{-\lambda h}=2-\cos (\alpha h)-\sqrt{(2-\cos (\alpha h))^{2}-1} e − λ h = 2 − cos ( α h ) − ( 2 − cos ( α h ) ) 2 − 1
Suppose that the random walk Z Z Z starts at position ( 0 , 1 ) ≡ ( 0 , N h ) (0,1) \equiv(0, N h) ( 0 , 1 ) ≡ ( 0 , N h ) at time 0 , and suppose that
τ = inf { n : Y n = 0 } . \tau=\inf \left\{n: Y_{n}=0\right\} . τ = inf { n : Y n = 0 } .
Stating fully any results to which you appeal, prove that
E exp ( i α X τ ) = e − λ E \exp \left(i \alpha X_{\tau}\right)=e^{-\lambda} E exp ( i α X τ ) = e − λ
where λ \lambda λ is as given at ( ∗ ) (*) ( ∗ ) . Deduce that as N → ∞ N \rightarrow \infty N → ∞
E exp ( i α X τ ) → e − ∣ α ∣ E \exp \left(i \alpha X_{\tau}\right) \rightarrow e^{-|\alpha|} E exp ( i α X τ ) → e − ∣ α ∣
and comment briefly on this result.