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Suppose that $X$ and $Y$ are independent random variables with the same geometric distribution, $\mathbb{P}(X=k)=\mathbb{P}(Y=k)=pq^{k-1}$ for $k\geq 1$, $q=1-p$.

Find $\mathbb{P}(X=k \mid X+Y=n+1)$, where $n$ is an element of $\left\{1,2,\dots\right\}$. What is this distribution?

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  • $\begingroup$ That should be $k \ge 1$, not $k > 1$. $\endgroup$ – Robert Israel Feb 17 '13 at 20:24
  • $\begingroup$ What did you try? (Removed irrelevant tag.) $\endgroup$ – Did Feb 17 '13 at 20:27
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    $\begingroup$ Hint: $P(A\mid B) = \frac{P(AB)}{P(B)}$. What is $P(AB)$ for your particular events $A$ and $B$, and can you express $P(AB)$ in terms of what the value of $X$ must be and what the value of $Y$ must in order for event $AB$ to occur? $\endgroup$ – Dilip Sarwate Feb 17 '13 at 20:30
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Hint: Consider repeatedly tossing a coin with probability $p$ of "heads" on each toss, independently. $X$ is the number of tosses until the first "heads" and $Y$ the number of tosses after that until the second "heads". So $X+Y=n+1$ says the second "heads" is on toss number $n+1$. Given that, the first "heads" could be on any of the first $n$ tosses...

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  • $\begingroup$ im sorry for commenting here, im having quite similar problem, to like this question. may i ask, why when computing $P(X+Y=n+1) = \sum_{i=1}^n \Pr(X=i)\Pr(Y=n+1-i)$? and suppose X is for trial before first head, and Y is trial after first head, and Z is trial when two heads appear, so Z=X+Y, what is we computer join distribution of X and Z? and how can i compute distribution of Z? $\endgroup$ – devss Jul 19 at 0:07
  • $\begingroup$ If $X$ and $Y$ are independent and take positive integer values, condition on the value of $X$. $\endgroup$ – Robert Israel Jul 19 at 3:59
  • $\begingroup$ thankyou, but $P(X,Y)$ means probability Y (head 2 times appear) and X (head first time appear)? while $P(X+Y)$ means probability Y (head 2 times appear) given that X (head first time appear)? $\endgroup$ – devss Jul 19 at 9:03
  • $\begingroup$ No, that all makes no sense. $X$ and $Y$ are random variables. $P(X)$ or $P(X,Y)$ or $P(X+Y)$ don't mean anything $P(X=x)$ means the probability that $X=x$ (i.e. $X$ takes the value $x$). $P(X=x, Y=y)$ is the probability that $X=x$ and $Y=y$. $P(X+Y=z)$ is the probability that the value of $X$ and the value of $Y$ sum to $z$. $\endgroup$ – Robert Israel Jul 19 at 13:39
  • $\begingroup$ ok thankyou! but im still confuse with the conditioning part in the $P(X=x, Y=y)$ ,$P(X+Y=z)$ means probability second heads is on toss number Z given that first toss is in X or it just means P(Z) which is Z is sum of X and Y? can i say $P(X+Y=z)$=$P(X=x).P(Z=z|X=x)$? or $P(X=x).P(Y=y|X=x)$?, $\endgroup$ – devss Jul 19 at 14:53
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If the answer is not intuitively obvious, we can always calculate. We want $\Pr(X=k|X+Y=n+1)$. By the usual expression for conditional probabilities, we have $$\Pr(X=k|X+Y=n+1)=\frac{\Pr((X=k) \cap (X+Y=n+1))}{\Pr(X+Y=n+1)}.$$

Calculate. The probability on top is the probability that $X=k$ and $Y=n+1-k$. This is $q^{k-1}p q^{n-k}p$, which simplifies to $p^2 q^{n-1}$.

The probability at the bottom is $\sum_{i=1}^n \Pr(X=i)\Pr(Y=n+1-i)$. This is $$\sum_{i=1}^n q^{i-1}p q^{n-i}p,$$ which is $np^2 q^{n-1}$.

Divide. We get $\dfrac{1}{n}$.

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