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I'm trying to solve this problem, but without much success. Here we have that $X \sim \operatorname{Geo}(\theta)$, with $p(x) = \theta (1 - \theta)^x$, and $Y = X \mod 3$, thereby having $p(y) = \theta\frac{ (1 - \theta)^y}{ 1 - (1 - \theta)^3}$, with $y = 0, 1, 2.$

Now I have to find the conditional distribution of $ Z = (X-Y)/3$, given $Y = y.$ I haven't found anything like this here (sorry if I duplicated a question, I tried not to). Can anyone help?

Maxim
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Bruna w
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    Note that $Y \in \{0,1,2\}$ . Given $Y$, you can easily calculate all possible values of $X$, and therefore the probability distribution of $X$ (just the normalized sum of the probabilities for the possible values of $X$.) After that, it should be obvious how to get to the conditional distribution of $Z|Y$. – jbowman Jan 10 '18 at 18:14
  • Thanks for the correction, but I still don't think it's that obvious ): – Bruna w Jan 10 '18 at 20:15
  • Please check that the corrections I made to your definitions and formulas conform with your intention. – whuber Jan 10 '18 at 20:32
  • @whuber how do you got this distribution for Y? – Bruna w Jan 10 '18 at 20:43
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    Let $y\in\{0,1,2\}$. Then $$\eqalign{\Pr(Y=y)&=\Pr(X\in\{y,y+3,\ldots,y+3n,\ldots\})\\&=\sum_{n=0}^\infty\theta(1-\theta)^{y+3n}\\&=\theta(1-\theta)^y\sum_{n=0}^\infty ((1-\theta)^3)^n\\&= \theta\frac{(1-\theta)^y}{1-(1-\theta)^3}.}$$(I accidentally dropped the factor of $\theta$ in my original edit.) – whuber Jan 10 '18 at 20:47
  • thanks! there is just a missing $\theta$ in the question. – Bruna w Jan 10 '18 at 20:50
  • nice! now, could you help me with the rest of the question, please? – Bruna w Jan 10 '18 at 20:51
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    Try working it out conditional upon $Y=0$. Once you've figured that out, the rest should be easier. What values can $X$ take when $Y=0$? And what is the probability distribution associated with those values? – jbowman Jan 10 '18 at 21:22
  • This sounds like a `self-study` tag is needed for this question. Please read the associated wiki and detail why you cannot solve the question. – Xi'an Jan 10 '18 at 21:31
  • I got P(Z = (X + Y)/3) = P(X = 3z + y) = $\theta (1 - \theta)^{3z + y}$. – Bruna w Jan 10 '18 at 22:40

1 Answers1

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Perhaps a solution based on understanding what geometric distributions mean would be of more interest than a purely algebraic one.

Preliminaries: notation; Geometric distributions

Recall that a Geometric distribution with parameter $\theta$ describes the chances of observing a sequence of $x\in\{0,1,2,\ldots\}$ failures before the first success in a series of independent Bernoulli$(\theta)$ trials, whose values I will write $U_1,U_2, \ldots, U_n,\ldots,$ encoding (as usual) $U_i=1$ to represent success. Writing $p_\theta(x)$ for those quantities, independence implies

$$p_\theta(x+1) = p_\theta(x)\Pr(U_{n+1}=1) = p_\theta(x)\theta.$$

Conversely, this relation completely determines the distribution from the facts that (1) all probabilities must sum to unity and (2) the nonzero probabilities are those for $x\in\{0,1,2,\ldots\}$.

Solution

Let's interpret the $Y$ and $Z$ of the problem. It is an obvious number-theoretic fact that any possible value $x$ of $X$ can be written in the form $$x=3z+y$$ where $y\in\{0,1,2\}$ and $z\in\{0,1,2,\ldots\}.$ When we condition on $Y=y$, we're saying there initially are $y$ failures and then there are $z$ groups of three failures each before success is observed. The independence of the Bernoulli trials in each group of three implies any sequence of three failures has a chance

$$\rho = (1-\theta)^3.$$

Consequently, the independence of each (nonoverlapping) group of three failures implies

$$\Pr(Z=z+1) = \Pr(Z=z) (1-\theta)^3= \Pr(Z=z) \rho$$

for any $z=0,1,2,\ldots.$ Thus, conditional on $Y=y$, $Z$ has a Geometric distribution with parameter $\rho$.

Among the salient implications of this observation--ones that immediately solve the problem--are

  1. The distribution of $Z$ is independent of $Y$.

  2. The distribution of $Z$ is Geometric with parameter $\rho=(1-\theta)^3.$

  3. You can write the probabilities down immediately using the usual formulas for the geometric distribution, using the parameter $\rho$.


For those who prefer pure algebra, an amusing (and perhaps surprising) solution method is provided by the technique of decimation described at https://stats.stackexchange.com/a/35138/919. This directly gives the distributions of $(Y,Z)$, from which the conditional distribution of $Z$ is found by dividing by the chance that $Y=y$.

whuber
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