Equality of distribution for random varibles

[logarithmic]

If 2 random variables X, and Y have the same distribution, does that mean that for another random variables Z, X + Z and Y + Z also have the same distribution?

From looking at the convolution formula, the answer should be yes, because the convolution of random variables depends only on the distribution of the 2 variables being added.

However, if we consider the Poisson process N(t), then the stationary increment property says that N(5) - N(4) has the same distribution as N(1).

Now consider the random variable (N(4) - N(5)) - N(1). If the above claim is true then this has the same distribution as N(1) - N(1), which has variance 0, i.e. Var((N(4) - N(5)) - N(1)) = 0. This would seem to imply that the count of random Poisson events between time 4 and 5 must always be exactly the same as between time 0 and 1 and there is actually no randomness at all, which can't possibly be true.

 

[LCKurtz]

If 2 random variables X, and Y have the same distribution, does that mean that for another random variables Z, X + Z and Y + Z also have the same distribution?

From looking at the convolution formula, the answer should be yes, because the convolution of random variables depends only on the distribution of the 2 variables being added.

However, if we consider the Poisson process N(t), then the stationary increment property says that N(5) - N(4) has the same distribution as N(1).

Now consider the random variable (N(4) - N(5)) - N(1)

But that isn't of the form N(r) - N(s).

 

[logarithmic]

But it equals N(1) - N(1) which is in the form N(r) - N(s).

What if we wrote it like this:

1. Clearly Var(N(1) - N(1)) = 0.
2. Now N(5) - N(4) has the same distribution as N(1) by the stationary increments property.
3. Thus (N(5) - N(4)) - N(1) has the same distribution as N(1) - N(1) (by the claim that if X has the same distribution as Y, then X - Z has the same distribution as Y - Z)
4. So, in particular, they have the same variance, i.e. Var((N(5) - N(4)) - N(1)) = Var(N(1) - N(1)) = 0.

 

[LCKurtz]

It has been a long time since I looked at this stuff, but I think your problem is that

N(5)-N(4) is itself not a member of your Poisson process N(t) and you can't claim the stationary increment property to (N(5)-N(4)) - N(1).

 

[logarithmic]

Hmm, I don't think I did that. The only time I used stationarity was at 2) where I said N(5) - N(4) has the same distribution as N(1). Which is OK. I didn't use stationarity to get (N(5) - N(4)) - N(1) having the same distribution as N(1) - N(1).