Calculate $\mathbb{E}[M_{\alpha}^{p}(t)]$ for all $p>0$ and $t>0$, where $M_{\alpha}(t):=e^{\alpha W_t-\frac{\alpha^2}{2}t}$, $t\ge 0$

86 Views Asked by At

I am going through this solved problem but I don't understand some steps. My professor is notorious for making errors very often so don't hold back if you think he's wrong... Or if I am wrong. I am going to write my questions in bold to make it easier. Thanks a lot in advance for any help you can provide!

EXERCISE: Let us consider the following martingale $$M_{\alpha}(t):=e^{\alpha W_t-\frac{\alpha^2}{2}t},$$ where $t\ge 0$, $\alpha\ge 0$.

A) Check that $\mathbb{E}M_{\alpha}(t)=1$.

B) Calculate $\mathbb{E}[M_{\alpha}^{p}(t)]$ for all $p>0$ and $t>0$.

ANSWER:

A) We observe that $M_{\alpha}(t)$ is a martingale. Hence, $$\mathbb{E}M_{\alpha}(t)=\mathbb{E}M_{\alpha}(0)=1$$ First Question: Why? This is what I get instead: $$\mathbb{E}M_{\alpha}(t)=\mathbb{E}\left(e^{\alpha W_t-\frac{\alpha^2}{2}t}\right)=\mathbb{E}\left(\frac{e^{\alpha W_t}}{e^{\frac{\alpha^2}{2}t}}\right)=\frac{e^0}{e^{\frac{\alpha^2}{2}t}}=\frac{1}{e^{\frac{\alpha^2}{2}t}}$$ $$\mathbb{E}M_{\alpha}(0)=\frac{e^0}{e^0}=1$$

B) It follows that

$$\mathbb{E}M_{\alpha}^{p}(t)=\mathbb{E}\left[M_{p\alpha}(t)e^{\frac{\alpha^2 p^2}{2}t-\frac{\alpha^2p}{2}t}\right]=e^{\frac{\alpha^2}{2}(p^2-p)t}$$ Second question: I don't understand the first equality above. In particular, why is this the result in the second equality below? $$\mathbb{E}M_{\alpha}^{p}(t)=\mathbb{E}\left(e^{\alpha W_t-\frac{\alpha^2}{2}t}\right)^p=\mathbb{E}\left[M_{p\alpha}(t)e^{\frac{\alpha^2 p^2}{2}t-\frac{\alpha^2p}{2}t}\right]=e^{\frac{\alpha^2}{2}(p^2-p)t}$$

1

There are 1 best solutions below

0
On BEST ANSWER

For reference what you have is an exponential martingale.

For A) I would just go about it that way $$ E[M_\alpha(t)]= e^{-\alpha^2/t}E[e^{\alpha W_t}]=e^{-\alpha^2/t}\int_Re^{\alpha \sqrt tZ} \phi(z) dz$$ where $\phi(.)$ is the normal pdf, compute the integral, done. B) is a generalization of the preceding steps.

Second question: $$E[M_\alpha(t)^p]= e^{-\alpha^2pt/2}E[e^{p\alpha\sqrt t Z}]=e^{-\alpha^2pt/2}e^{p^2\alpha^2t/2}=e^{\alpha^2t/2(p^2-p)}$$ the second equality is motivated by the definition of the MGF for a std normal variable