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Consider three random variables X, Y, and Z, associated with the same experiment. The random variable X is geometric with parameter p∈(0,1). If X is even, then Y and Z are equal to zero. If X is odd, (Y,Z) is uniformly distributed on the set S={(0,0),(0,2),(2,0),(2,2)}. The figure below shows all the possible values for the triple (X,Y,Z) that have X≤8. (Note that the X axis starts at 1 and that a complete figure would extend indefinitely to the right.)

Joint PMF calculation

Use x, p express:

1) If x is odd and (y,z)∈{(0,0),(0,2),(2,0),(2,2)}, pX,Y,Z(x,y,z)=

2) If x is even and (y,z)=(0,0), pX,Y,Z(x,y,z)=

3) If x is odd, pX,Y(x,2)=

4) pY(2)=

5) var(Y+Z∣X=5)=

I can't really understand this question, when X is geometric with parameter p∈(0,1), how can I join it with (Y,Z)? Doesn't it mean X is odd or even with p and 1-p?

Q Yang
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  • There are two closely related but inequivalent definitions of #X# geometric with parameter $p$. The more common is that you repeat independently an experiment with probability of success $p$ and of failure $1-p$ until the first success. $X$ is the number of trials we use. We have $\Pr(X=k)=(1-p)^{k-1}p$. – André Nicolas Mar 09 '15 at 02:57
  • I understand the definition of geometric random variable, but I really don't know how to use it to calculate joint PMF. Could you give me an example for the first question? Maybe I can get hint on how to apply this definition in actual calculation. – Q Yang Mar 09 '15 at 03:08
  • You can calculate the probability that $X$ is odd (so the probability that $X$ is even. The probability that $X$ is odd is the probability it is $1$ plus the probability it is $3$, plus the probability it is $5$, and so on. This is $p+(1-p)^2p +(1-p)^4p+\cdots$, a geometric series whose sum can be found. – André Nicolas Mar 09 '15 at 05:09
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    @ André Nicolas. Thanks for hint, I figured out P(X is odd)=1/(2-p) and P(X is even)=(1-p)/(2-p).Great, so the answer for part 4 is (1/2)P(X is odd). – Q Yang Mar 10 '15 at 05:59

1 Answers1

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1) If $x$ is odd and $(y,z)∈\{(0,0),(0,2),(2,0),(2,2)\}$, $p_{X,Y,Z}(x,y,z)=$

We are given the distribution of $(Y,Z)$ when $X$ is odd.   It's uniform over four points.

Next, if $X$ is odd then the conditional probability of $X=x$ (where $x$ is a particular odd value) will be the probability of a success occurring on the $(x+1)/2$ -th odd value (given it does not occur on any even ones). Thus the conditioned value is Geometrically Distributed.

$$\begin{align} p_{_{Y,Z}}(y,z\mid \operatorname{Odd}(X)) & = \frac 1 4 \;\mathbf 1_{(y,z)\in \{(0,0),(0,2),(2,0),(2,2)\}} \\[2ex] p_{_X}(x\mid \operatorname{Odd}(X)) & = p(1-p)^{(x-1)/2} \mathbf 1_{x\in \Bbb Z^+\setminus \Bbb 2Z} \\[2ex] p_{X,Y,Z}(x,y,z\mid \operatorname{Odd}(X)) & =p_{_X}(x\mid \operatorname{Odd}(X)) \; p_{_{Y,Z}}(y,z\mid \operatorname{Odd}(X)) \\[1ex] & = \frac 1 4 p(1-p)^{(x-1)/2} \mathbf 1_{x\in \Bbb Z^+\setminus \Bbb 2Z}\;\mathbf 1_{(y,z)\in \{(0,0),(0,2),(2,0),(2,2)\}} \end{align}$$

2) If $X$ is even and $(Y,Z)=(0,0)$, $p_{X,Y,Z}(x,y,z)=$

We are given the distribution of $(Y,Z)$ when $X$ is even; it's a certainty at one point.

$$p_{Y,Z}(y,z \mid \operatorname{Even}(X)) = \mathbf 1_{y=0, z=0}$$

All we need is the conditional distribution of $X$ given that it is even.

As before we argue that if $N= X/2 \mid_{X\in 2\Bbb Z^+}$ then $N$ will be geometrically distributed.

$$p_X(x\mid \operatorname{Even}(X)) = p(1-p)^{x/2-1}$$

Put these together.

3) If $X$ is odd, $p_{X,Y}(x,2\mid \operatorname{Odd}(X))=$

Similar.

4) $p_Y(2)=$

$p_Y(2\mid \operatorname{Odd}(X)) = \frac 1 2 \\[2ex] p_Y(2) = p_Y(2\mid \operatorname{Odd}(x)) \; \mathsf P(\operatorname{Odd}(x)) $

Now, what is: $\operatorname P(\operatorname{Odd}(X))\;$?

5) $\mathsf{Var}(Y+Z\mid X=5)=$

We know $\mathsf P_{Y,Z}(y,z\mid X=5) = \frac 1 4 \;\mathbf 1_{(y,z)\in \{(0,0), (0,2), (2,0), (2,2)\}}$ so then:

$$\mathsf{Var}(Y+Z\mid X=5)= \frac 1 {\color{red}{4}}(0+4+4+16) - \frac 1{16}(0+2+2+4)^2 \\ = 2$$

Graham Kemp
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  • For part 2, given if X is even, then Y and Z are equal to zero. Sounds like pX,Y,Z(x,y,z)=1. However, I am not sure whether Y=Z=0 means (y,z)=(0,0)? If not, pX,Y,Z(x,y,z)=0. Could you explain it a little more? Thanks a lot. – Q Yang Mar 09 '15 at 04:21
  • @QYang Yes, $;p_{{Y,Z}}(y,z \mid \operatorname{Even}(X)) = \mathbf 1{y=0, z=0};$, but $$;p_{{X,Y,Z}}(x,y,z\mid \operatorname{Even}(X)) = p{{Y,Z}}(y,z \mid \operatorname{Even}(X)); p{_{X}}(x\mid\operatorname{Even}(X));$$ What is the conditional probability that $X$ will take a particular even value, when it is given that it will take some even value? – Graham Kemp Mar 09 '15 at 04:37
  • OK, I see. I didn't join them. P(even)=p*(1-p)^(x/2-1). There is another question, I assume P(even)=1-P(odd), is it right? BTW, I didn't work out part 5 yet, var=-2.5 in your solution, . But consider var>=0, it could be a wrong answer. ;) – Q Yang Mar 09 '15 at 04:50
  • @QYang Typo. Can you see from where the figures come? – Graham Kemp Mar 09 '15 at 05:05
  • @ Graham Kemp which figures we are talking about? the var result or the original number provided in question? – Q Yang Mar 09 '15 at 05:09
  • @ Graham Kemp. Yes, I am just wondering why E[X^2] and (E[X])^2 both need to square the Py,z. Do you think part 3 and part 4 are the same questions? Because pY(2∣even(x))=0, so pX,Y(x,2∣odd(x))=pY(2). – Q Yang Mar 09 '15 at 05:24
  • @ Graham Kemp. Finally, I figured out part 3 and part 4 are not same, for part 4, I should sum P(odd(x)).... After submitting the homework, it showed the first four answers were wrong. But it doesn't matter, thanks for your patient explanation. I have the better understanding about how joint PMF and geometric RV work. When final result come out, I will post the solution here. – Q Yang Mar 09 '15 at 06:30
  • For the first and second part, we cannot change the power values on (1-p) based on odd or even since they are the random variables and not some unrelated counts. – bicepjai Mar 03 '17 at 01:39