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

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.

• What is the $G$ distribution? May 19 '19 at 10:42
• @IshanDeo geometric distribution May 19 '19 at 10:42

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.