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Let $X$ and $Y$ be independent random variables. Each of them has a geometric distribution with $E[X] = 2$ and $E[Y] = 3$.

(a) Find the joint p.m.f. of $X$ and $Y$.

(b) Compute the probability that $X + Y \le 4$.

(c) Define two new random variables by $W = \min{ \left( X, Y \right ) }$ and $Z = \max{ \left( X, Y \right ) }$. Find the joint p.m.f. of $W$ and $Z$.

Ans:

$X$ and $Y$ are independent random variables. Each of them has a geometric distribution with $E[X] = 2$ and $E[Y] = 3$.

So joint p.m.f. of $X$ and $Y$,

$E[XY] = E[X] E[Y] = 2 * 3 = 6$

Is this correct?

And for $(b)$ and $(c)$ can anyone suggest how to proceed??

for (c), s min(X; Y ) is a geometric random variable with parameter 1 - (1 - p)^2 = 2p - p^2

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    $\begingroup$ Sorry but do you know what is a PMF? $\endgroup$
    – Did
    Oct 1, 2013 at 15:44
  • $\begingroup$ Probability mass function $\endgroup$
    – nini
    Oct 1, 2013 at 15:49
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    $\begingroup$ Then why none appear in your question? In particular the passage "So joint p.m.f. of X and Y, E[XY ] = E[X] E[Y ] = 2 *3 = 6 Is this correct?" seems quite misleading. $\endgroup$
    – Did
    Oct 1, 2013 at 15:52
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    $\begingroup$ Sure, if you tell us what is a probability mass function... $\endgroup$
    – Did
    Oct 1, 2013 at 16:38
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    $\begingroup$ And when they say that X has a geometric distribution with E[X]=2 this means that P(X=x) is what? $\endgroup$
    – Did
    Oct 1, 2013 at 18:40

2 Answers 2

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Hint
The pmf for geometric distribution (the probability that it will take $k$ trials to get the first success) is given by $$\Pr(X = k) = (1-p)^{k-1}\,p\,$$ where $p$ is the success probability in each trial.
It is also known that the mean of $X$ is $\mathrm{E}(X) = \frac{1}{p}$.
Can you get it from here?

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  • $\begingroup$ Thanks for (a) So u suggest pmf is E(X) = 2 ???... Any help on (b) and (c), can you help? $\endgroup$
    – nini
    Oct 1, 2013 at 16:58
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    $\begingroup$ @nini no way. I didn't suggest that. pmf is a function of trials ($k$) and you say it is a constant. $\endgroup$ Oct 1, 2013 at 17:08
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    $\begingroup$ @nini so then you almost got (a). Just remember that they are independet $\endgroup$ Oct 1, 2013 at 17:13
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    $\begingroup$ @nini for (c) you have to remember that $p$ is different $1/2$ and $1/3$ (do you see the product sign over there where you have found that formula?): $p = 1-\prod_m(1-p_{m})$. In your case $p_1=1/2$ and $p_2=1/3$ so $p=2/3$. $\endgroup$ Oct 1, 2013 at 17:34
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    $\begingroup$ @nini guess this can be helpful. $\endgroup$ Oct 1, 2013 at 18:01
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$P=1/E[X]=0.5$

$P=1/E[y]=0.333$

$PX(x)=(1-0.5)^{k-1}0.5$

$PY(y)=(1-0.33)^{k-1}0.333$

The Joint Probability:

$E[XY]=E[X]*E[y]=2\cdot3=6$

$p=1/E[XY]=0.167$ $PX,Y(x,y)=(1-0.167)^{k-1}0.167 ~~~ K=1,2,3.......$

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