Stochastic calculus:Laplace transformation of a Wiener process

[the_dane]

Homework Statement

I am asked to show that $$E[\exp(a*W_t)]=\exp(\frac{a^2t}{2})$$
Let's define: $$Z_t = \exp(a*W_t)$$
W_t is a wiener process

Homework Equations

$$W_t \sim N(0,\sqrt{t})$$

The Attempt at a Solution

I want to use the following formula.
if Y has density f_Y and there's a ral function g then the following holds:
$$E[g(Y)] = \int_{-\infty}^{\infty} g(u)f_Y(u)du$$
In my Case:
$$E(Z_t)= \int_{-\infty}^{\infty} \exp(a*u) \frac{1}{\sqrt{2\pi \sqrt{t}}} \exp(-(u)^2/ 2\sqrt{t})du$$
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi \sqrt{t}}} \exp(a*u-(u)^2/ 2\sqrt{t})du$$.
Then I notice; IF $$= \exp(a*u-(u)^2/s\sqrt{t}) = \exp(-(u-\exp(\frac{a^2t}{2}))^2/ 2\sqrt{t})$$.
Then Z_t must be normally distributed with mean $$\exp(\frac{a^2t}{2})$$
Unfortunately my creativity ends here and I cannot show the last part. It should "just" simple "moving terms around" though, right?

Please confirm that my approach is correct and please help me finish. thx

 

[Ray Vickson]

Homework Statement

I am asked to show that $$E[\exp(a*W_t)]=\exp(\frac{a^2t}{2})$$
Let's define: $$Z_t = \exp(a*W_t)$$
W_t is a wiener process

Homework Equations

$$W_t \sim N(0,\sqrt{t})$$

The Attempt at a Solution

I want to use the following formula.
if Y has density f_Y and there's a ral function g then the following holds:
$$E[g(Y)] = \int_{-\infty}^{\infty} g(u)f_Y(u)du$$
In my Case:
$$E(Z_t)= \int_{-\infty}^{\infty} \exp(a*u) \frac{1}{\sqrt{2\pi \sqrt{t}}} \exp(-(u)^2/ 2\sqrt{t})du$$
$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi \sqrt{t}}} \exp(a*u-(u)^2/ 2\sqrt{t})du$$.
Then I notice; IF $$= \exp(a*u-(u)^2/s\sqrt{t}) = \exp(-(u-\exp(\frac{a^2t}{2}))^2/ 2\sqrt{t})$$.
Then Z_t must be normally distributed with mean $$\exp(\frac{a^2t}{2})$$
Unfortunately my creativity ends here and I cannot show the last part. It should "just" simple "moving terms around" though, right?

Please confirm that my approach is correct and please help me finish. thx

This just involves the moment-generating function of the normal distribution. It does not really use any deep results about Wiener processes, just the fact that $W_t \sim \text{N}(0,\sqrt{t})$.

See, eg., https://www.le.ac.uk/users/dsgp1/COURSES/MATHSTAT/6normgf.pdf

 

[Orodruin]

You did not get the Wiener process correctly, it is $W_t \sim \mathcal N (0,t)$, not $\mathcal N(0,\sqrt{t})$, i.e., the variance is $t$, not $\sqrt t$. What is $\sqrt t$ is the standard deviation. This propagates to your expressions for the pdf.

Once this is corrected, it should just be a matter of completing the square in the argument of the exponent, changing variables in the integral, and performing the resulting Gaussian integral.

 

[Ray Vickson]

You did not get the Wiener process correctly, it is $W_t \sim \mathcal N (0,t)$, not $\mathcal N(0,\sqrt{t})$, i.e., the variance is $t$, not $\sqrt t$. What is $\sqrt t$ is the standard deviation. This propagates to your expressions for the pdf.

Once this is corrected, it should just be a matter of completing the square in the argument of the exponent, changing variables in the integral, and performing the resulting Gaussian integral.

Some sources use the notation $N(\mu,\sigma)$ while others use $N(\mu,\sigma^2)$. As long as the OP uses notation consistent with his/her textbook or course notes, that is sufficient.

 

[Orodruin]

As long as the OP uses notation consistent with his/her textbook or course notes, that is sufficient.

I agree, but this is not the case. The pdf used by the OP has variance $\sqrt{t}$, not $t$.

Edit: Of course, all this will do is to change the $t$ in the correct answer to a $\sqrt t$. More generally, the expectation of $\exp(aX)$ where $X\sim \mathcal N(0,\sigma^2)$ would be $\exp(a^2\sigma^2/2)$. Regardless, the way forward is the same. Complete the square, change variables, integrate.

 

[Ray Vickson]

I agree, but this is not the case. The pdf used by the OP has variance $\sqrt{t}$, not $t$.

OK: I did not notice that. Well spotted.

Edit: Of course, all this will do is to change the $t$ in the correct answer to a $\sqrt t$. More generally, the expectation of $\exp(aX)$ where $X\sim \mathcal N(0,\sigma^2)$ would be $\exp(a^2\sigma^2/2)$. Regardless, the way forward is the same. Complete the square, change variables, integrate.