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Let $Z \sim G(p), W \sim G(2p)$ be independent random variables so that $ P (W>Z-1) = \frac{3}{7}$. Calculate $p$.

I tried to solve it this way: $$ P (W>Z-1) = \sum_{k=1}^{\infty} P(W-k+1>0, Z =k) $$ and then used the assumption that they are independent and have geometric distribution, but few steps further I got stuck and I don't know what other way I can solve this.

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  • $\begingroup$ What is the $G$ distribution? $\endgroup$
    – Ishan Deo
    May 19 '19 at 10:42
  • $\begingroup$ @IshanDeo geometric distribution $\endgroup$
    – user560461
    May 19 '19 at 10:42
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For a random variable $X \sim G(p)$ with probability mass function given as, $P(X = k) = p(1-p)^{k-1}, \, k = 1,2,3...$

It can be easily shown that, $P(X > x) = (1-p)^{x}, \, x = 0,1,2...$

Now consider $Z \sim G(p)$ and $W \sim G(2p)$, such that $Z$ and $W$ are independent. Then,

$P(W > Z-1 \mid Z = z) = (1-2p)^{z-1}$

$P(W > Z-1 \cap Z = z) = P(W > Z-1\mid Z = z).P(Z = z) = (1-2p)^{z-1}.p(1-p)^{z-1}$

$P(W > Z-1) = \displaystyle \sum_{z=1}^{\infty}P(W > Z-1 \cap Z = z) = \displaystyle \sum_{z=1}^{\infty}(1-2p)^{z-1}.p(1-p)^{z-1}$

$P(W > Z-1) = p\displaystyle \sum_{z=1}^{\infty}\left((1-p)(1-2p)\right)^{z-1}$

$P(W > Z-1) = \dfrac{p}{(1-p)(1-2p)} = \dfrac{3}{7}$

So, you get a quadratic expression in $p$ which can be easily solved.

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