Expected value of stationary independent increments.

[dionysian]

Homework Statement

Let $$\{ X(t),t \ge 0\}$$ be a random process with stationary independent increments and assume that $$X(0)=0$$. Show that:

$$E[X(t)] = {\mu _1}t$$

Homework Equations

The Attempt at a Solution

I tried to work backwards by argueing that the mean between time intervals it equal to the product of the mean at t =1 and the time interval
$$\mu \Delta {t_1} + \mu \Delta {t_2} + \cdot \cdot \cdot + \mu \Delta {t_n} = \mu (\Delta {t_1} + \Delta {t_2} + \cdot \cdot \cdot + \Delta {t_n}) = \mu {t_n}$$

Allthough this gives me the write ansewr i don't feel its a very solid proof. Its seems there should be a straight forward way to show this with the diffenition of the expected value but i can't seem to see it.

 

[mighty2000]

Thank you for your question. I will provide a proof for the statement using the definition of expected value.

First, we know that a random process with stationary independent increments means that the increments of the process in non-overlapping time intervals are independent of each other and have the same distribution. This means that for any time interval [t1,t2], the increment X(t2) - X(t1) has the same distribution as X(t2') - X(t1') for any other non-overlapping interval [t1',t2'].

Next, we can define the expected value of a random process as follows:

E[X(t)] = ∫x f(x) dx

where f(x) is the probability density function of the process at time t and x represents the possible values of the process at time t.

Since we are assuming that X(0) = 0, we can rewrite this as:

E[X(t)] = ∫x f(x) dx = ∫x f(x) dx + ∫0 f(0) dx

= ∫x f(x) dx + 0

= ∫x f(x) dx

Now, since the increments of the process are independent, we can rewrite the integral as:

∫x f(x) dx = ∫x (f(x1) + f(x2) + ... + f(xn)) dx

where x1, x2, ... , xn are the possible values of the process in non-overlapping time intervals [t1,t2], [t2,t3], ... , [tn-1,tn].

Since the increments have the same distribution, we can also rewrite the integral as:

∫x (f(x1) + f(x2) + ... + f(xn)) dx = ∫x (f(x) + f(x) + ... + f(x)) dx

= ∫x nf(x) dx

= n∫x f(x) dx

= nE[X(1)]

Therefore, we have shown that:

E[X(t)] = ∫x f(x) dx = nE[X(1)]

= n(∫x f(x) dx)

= nE[X(1)] = n(μ1)

= μ1t

This proves that E[X(t)] = μ1t, as desired.

I hope this helps. Let me know if you have any further questions