Probability distributions

[squenshl]

Homework Statement

Let $$\Omega$$ = {w₁, w₂, w₃}, P(w₁) = 1/3, P(w₂) = 1/3, P(w₃) = 1/3, and define X, Y, Z as follows:
X(w₁) = 1, X(w₂) = 2, X(w₃) = 3
Y(w₁) = 2, Y(w₂) = 3, Y(w₃) = 1
Z(w₁) = 3, Z(w₂) = 1, Z(w₃) = 2

(a) Show that these 3 random variables have the same distribution.
(b) Find the probaility distribution of X+Y, Y+Z and X+Z.
(c) Show that X and Y are not independent by verifying that Bienaymé's formula doesn't hold.
(d) Find a random variable W such that X and W are independent.

Homework Equations

The Attempt at a Solution

For (a) do we just add the values of X(w₁), X(w₂) and X(w₃) with the values of Y(w₁), Y(w₂) and Y(w₃) and so on and so forth. Find out they add to the same values so they must have the same prob distn.

 

[squenshl]

Or is it for:
X: 1 x 1/3 + 2 x 1/3 + 3 x 1/3 = 2
Y: 2 x 1/3 + 3 x 1/3 + 1 x 1/3 = 2
Z: 3 x 1/3 + 1 x 1/3 + 2 x 1/3 = 2

Since the 3 random variables have the same expectation, they have the same probability distribution.

 

[squenshl]

I have done (a), (b), (c) but how do I do (d). I was told to show that P(X $$\cap$$ W) = P(X)P(W) but how do I do that?

 

[Gregg]

Since the 3 random variables have the same expectation, they have the same probability distribution.

3 random variables with the same expectation do not necessarily have the same distribution. The expectation is just the mean, you can find a counter example to that easily. MGFs can show that distributions are the same?

 

[Ray Vickson]

Homework Statement

Let $$\Omega$$ = {w₁, w₂, w₃}, P(w₁) = 1/3, P(w₂) = 1/3, P(w₃) = 1/3, and define X, Y, Z as follows:
X(w₁) = 1, X(w₂) = 2, X(w₃) = 3
Y(w₁) = 2, Y(w₂) = 3, Y(w₃) = 1
Z(w₁) = 3, Z(w₂) = 1, Z(w₃) = 2

(a) Show that these 3 random variables have the same distribution.
(b) Find the probaility distribution of X+Y, Y+Z and X+Z.
(c) Show that X and Y are not independent by verifying that Bienaymé's formula doesn't hold.
(d) Find a random variable W such that X and W are independent.

Homework Equations

The Attempt at a Solution

For (a) do we just add the values of X(w₁), X(w₂) and X(w₃) with the values of Y(w₁), Y(w₂) and Y(w₃) and so on and so forth. Find out they add to the same values so they must have the same prob distn.

(a) P{X=1} = P(w_1) = 1/3, P{Y=1} = P(w_3) = 1/3, etc. So, they all have the same distribution.

(b) P{X+Y=2} = P{X=1 &Y=1} = P{w_1 & w_2} = 0, P{X+Y=3} + P{X=1 & Y=2} + P{X=2 & Y=1} = P{w_1 & w_1} + P{w_2 & w_3} = P{w_1} = 1/3. P{X+Y=4} = P{w_2 & w_2} = 1/3, and P{X+Y = 6} = P{w_3 & w_3} = 1/3, so the probabilities P{X+Y= = i} for i=2,3,4,5,6 are 1/3,0,1/3,0,1/3. You can do Y+Z and X+Z similarly.

(c) P{X=1 & Y=1} = 0, while P{X=1}*P{Y=1} = (1/3)^2 = 1/9.

(d) An obvious solution W = k (k = an arbitrary constant) for all three points w_i . This is a degenerate random variable. With more work (eg., using moment-generating functions) we can show that this is the _only_ solution.

R.G. Vickson