X and Y are both geometric distributions with success p. What it the Pr(X+Y=n)
Would I use a convolution with a sum for this? Do I need to define a third random variable?
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1Are $X$ and $Y$ independent? – Clarinetist Jun 23 '16 at 11:54
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yep forgot to mention they are independant variables – RonaldB Jun 23 '16 at 12:00
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1Do you know about MGFs and independence with MGFs? – Clarinetist Jun 23 '16 at 12:01
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@Clarinetist no we haven't learned those. – RonaldB Jun 23 '16 at 12:01
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1Convolution would be the way to go, then, I would think. – Clarinetist Jun 23 '16 at 12:02
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@Clarinetist Do I need to define another random variable i.e. X=Z-Y? – RonaldB Jun 23 '16 at 12:05
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2Something like that. Another question: there are two types of geometric distributions. One is the number of trials until the first success, and the other is the number of failures before the first success. Which one are you working with here? – Clarinetist Jun 23 '16 at 12:07
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@Clarinetist number of trials until first success – RonaldB Jun 23 '16 at 12:10
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1See https://en.wikipedia.org/wiki/Negative_binomial_distribution – Henry Jun 23 '16 at 12:44
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From the definition of Geometric Random Variables:
If $X, Y$ are the independent counts of trials, from sequences of Bernoulli trials, until first success (with rate $p$), then $X+Y$ would be the count of trials from a sequence of Bernoulli trials until the second success. We would then be seeking the probability of obtaining exactly one success somewhere among $n-1$ trials, then a second success.
$$\mathsf P(X+Y=n) = (n-1) p^2 (1-p)^{n-2}~\mathbf 1_{n\in \{2,3,\ldots\}}$$
Graham Kemp
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We have \begin{align} &\ \Pr(X + Y = n)\\ = &\ \sum_{i=1}^{n-1} \Pr(X=i) \cdot \Pr(Y = n - i \mid X = i) \\ = &\ \sum_{i=1}^{n-1}\Pr(X = i)\cdot\Pr(Y=n-i) \\ = &\ \sum_{i=1}^{n-1}p(1-p)^{i-1} \cdot p(1-p)^{n-i-1} \\ = &\ \sum_{i=1}^{n-1} p^2(1-p)^{n-2} \\ = &(n-1)p^2(1-p)^{n-2} \end{align}
PSPACEhard
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