Solving Stochastic Processes Homework Problem

[wildman]

Homework Statement

I need someone to reassure me (or correct me) on this problem:

The process $$X(t) = e^{At}$$ is a family of exponentials depending on the random variable A.
Express the mean $$\eta(t)$$, the autocorrelation $$R(t_1,t_2)$$, and the first order density f(x,t) of X(t) in terms of the density $$f_a(a) of A$$

Homework Equations

$$f(x,t) = \frac {\partial F(x,t)} {\partial x}$$
$$\eta (t) = \int_{-\infty}^{\infty} xf(x,t)dx$$
$$R(t_1,t_2) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x_1 x_2$$
$$f(x_1,x_2;t_1,t_2)dx_1 dx_2$$

The Attempt at a Solution

$$f_a(a) = ae^a$$
$$f(x,t) = ae^{at}$$
$$\eta(t) = \int_{-\infty}^{\infty} a^2 e^{at} da$$
$$R(t_1,t_2) = \int_{-\infty}^{\infty} a^4 e^{at_1} e^{at_2} da$$

 

[mighty2000]

Firstly, your attempt at a solution is correct. The mean, autocorrelation, and first order density of X(t) can all be expressed in terms of the density of A.

To further reassure you, let's break down the problem and explain why this is the case.

The process X(t) = e^{At} is a continuous stochastic process, meaning it is a function of both time and a random variable A. This random variable A has a probability density function (PDF) f_a(a), which describes the distribution of possible values for A.

Now, the mean \eta(t) of X(t) is simply the expected value of X(t) at any given time t. We can express this in terms of the PDF of A by using the formula \eta(t) = \int_{-\infty}^{\infty} xf(x,t)dx, where f(x,t) is the joint PDF of X(t) and A. Since X(t) = e^{At}, we can substitute this into the formula and get \eta(t) = \int_{-\infty}^{\infty} ae^{at}f_a(a)da. This is the same as your solution, \int_{-\infty}^{\infty} a^2 e^{at} da.

Similarly, the autocorrelation R(t_1,t_2) is a measure of the correlation between two different time points of X(t). It can be expressed as \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x_1 x_2 f(x_1,x_2;t_1,t_2)dx_1 dx_2, where f(x_1,x_2;t_1,t_2) is the joint PDF of X(t_1) and X(t_2). Again, since X(t) = e^{At}, we can substitute this into the formula and get R(t_1,t_2) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{at_1} e^{at_2} f_a(a)da. This is the same as your solution, \int_{-\infty}^{\infty} a^4 e^{at_1} e^{at_2} da.

Finally, the