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I'm really struggling to figure this question out.

Let $Y, Z$ and $W$ be independent, identically distributed $Pois(\lambda)$ random variables. Let \begin{align}X1&=Y\\ X2 &= Y − Z\\X3 &= Y + Z + W.\end{align}

(d) Are $X_1$ and $X_3$ independent? Give justification for your answer.

(e) Let $p_{X_1,X_3}$ be the joint probability mass function of $X_1$ and $X_3$. Calculate $p_{X_1,X_3}(0,0)$.

So far I managed to prove independence since Cov is not equal to zero so they are not independent and they both involve $Y$ but can't seem to figure out how to do (e).

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    $\begingroup$ The only way $X_1=X_3=0$ is possible is if $Y=0$ and $Y+Z+W=0$, which is to say... $\endgroup$ Aug 12 '17 at 18:12
  • $\begingroup$ does this Z and W are zero but surely Z and W can be W= -Z for any positive number Z $\endgroup$ Aug 12 '17 at 19:57
  • $\begingroup$ What is the chance that a $\operatorname{Pois}(\lambda)$ ranom variable like $W$ can equal -3, say? $\endgroup$ Aug 12 '17 at 19:59
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    $\begingroup$ that is what i thought so does that mean X=Y=Z=0 so the answer is e^(-3lambda) right? $\endgroup$ Aug 12 '17 at 20:09
  • $\begingroup$ That's what I get. $\endgroup$ Aug 12 '17 at 21:47
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$Y \sim \mathsf{Pois}(\lambda),$ so $P(Y = 0) = e^{-\lambda}.$

Let $U = Z + W \sim \mathsf{Pois}(2\lambda),$ because $Z$ and $W$ are independent. Then $P(U = 0) = e^{-2\lambda}.$

Now $p_{X_1, X_3}(0,0) = P(Y = 0, U = 0) = e^{-\lambda} \times e^{-2\lambda} = e^{-3\lambda}.$

A Poisson distribution does not take negative values, so there is no need to consider that possibility.


Mainly for visualization, here is a simulation of the relevant distributions with $\lambda = 2.$ With a million iterations of the experiment, answers should be accurate to two or three significant digits.

m = 10^6;  lam = 2
y = rpois(m,lam); z = rpois(m,lam);  w = rpois(m,lam)
x1 = y;  x2 = y-z;  x3 = y+z+w
mean(x1);  mean(x2);  mean(x3)
## 2.000591  # aprx E(X1) = 2
## 0.001449  # aprx E(X2) = 0
## 5.99855   # aprx E(X3) = 6
var(x1);  var(x2);  var(x3)
## 2.001833  # aprx Var(X1) = 2
## 3.995877  # aprx Var(X2) = 4
## 5.996642  # aprx Var(X3) = 6
cov(x1, x3);  mean(x1==0 & x3 == 0);  exp(-3*lam)
## 2.003265  # aprx Cov(X1, X3) = Var(X1) = 2
## 0.002478  # aprx P(X1 = 0, X3 = 0)
## 0.002478752 # exact P(X1 = 0, X3 = 0)

The figure below gives a rough visualization of the joint distribution of $X_1$ and $X_3.$ Actually all points in each rectangle fall at the center of the rectangle (integer values), but they have been 'jittered' (randomly displaced) to prevent massive overplotting. For a clearer graph only 10,000 simulated points are shown.

enter image description here

Notice in particular the very few points (indicating low probability, about 0.0025) in the rectangle at $(0, 0).$ Also, it is clear from the plot that $X_1$ and $X_2$ are not independent.

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