I havent been able to find an analogous question and our textbook is lacking in good examples, so I could use a little help with this rather straight forward martingale problem:
Let X=(Xn) be a sequence of independent random variables on (omega,ℙ,F) where ℙ(Xn=1) = p and ℙ(Xn=-1) = q := 1-p Set Yn = ΣXi for i to n Define Mn = (e^(aYn))(pe^a + qe^-a)^-n for a>0 Show that Mn is a Fn-martingale
So I think I understand the idea of the martingale and the objective of the proof. A martingale is a process where ExpectedValue(Mn+1) = Mn. So to prove something is a martingale we can prove that or equivalently that ExpectedValue(Mn+1) - ExpectedValue (Mn) = 0.
I started by saying that we know: Mn+1 = Mn(e^(aYn+1))(pe^a + qe^-a)^(-n+1) From there I take the expectation of each side: ExpectedValue(Mn+1|Fn) = ExpectedValue(Mn(e^(aYn+1))(pe^a + qe^-a)^(-n+1)) We can bring the Mn out because ExpectedValue(Mn) = Mn: = Mn(ExpectedValue((e^(aYn+1))(pe^a + qe^-a)^(-n+1)))
But I get stuck as this point and I am not sure what the next step is (or if I went awry earlier). It seems like everything left in the expected value should but the exponential seems problematic in that respect. Perhaps I am going about it wrong?
Thanks in advance for your time and help!
You need to prove $$ E(M_n | \mathcal F_{n-1}) = M_{n-1} $$ where $\mathcal F_n$ is the minimal sigma algebra with respect to which $X_1,\dots,X_n$ are measurable.
Now $M_n = e^{aY_n}(pe^a + qe^{-a})^{-n}$. Let's compute: $$ E(M_n | \mathcal F_{n-1}) = p \cdot e^{a(Y_{n-1} + 1)}(pe^a + qe^{-a})^{-n} + q \cdot e^{a(Y_{n-1} - 1)}(pe^a + qe^{-a})^{-n}) ,$$ since $Y_{n-1}$ is $\mathcal F_{n-1}$ measurable. After this it is just algebra.