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Your first success probability $p_a$ is higher on the first trial, than on the remaining trials (where it is $p_b$, constantly). How does this impact the mean of the general experiment ($1/p$ in the general, fixed geometric case) and the standard deviation ($\sqrt{(1-p)/p^2}$ in the general, fixed geometric case)?

EDIT: I think I got the mean:

$p_a + (1-p_a)(1/p_b+1)$

But I'm still unsure about the stdev

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1 Answer 1

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It seems you've already figured out how to adjust the expected value: Consider the experiment as one trial with probability $p_a$, and with probability $1-p_a$ a standard series of Bernoulli trials with parameter $p_b$ plus one extra trial.

Since the variance can be written as the difference of two expectation values, you can do the same thing for the variance. From $\langle n\rangle=1/p$ and $\sigma^2=(1-p)/p^2=\langle n^2\rangle-\langle n\rangle^2$ in the unmodified case, we get $\langle n^2\rangle=(1-p)/p^2+1/p^2=(2-p)/p^2$. Thus in the modified case, we have

$$ \begin{align} \langle n'\rangle&=p_a+(1-p_a)\langle n+1\rangle\\ &=p_a+(1-p_a)(1/p_b+1)\\ &=1+(1-p_a)/p_b\;,\\ \langle n'-1\rangle^2&=\left(\langle n'\rangle-1\right)^2\\ &=(1-p_a)^2/p_b^2\;,\\ \langle(n'-1)^2\rangle&=p_a\cdot0+(1-p_a)\langle n^2\rangle\\ &=(1-p_a)(2-p_b)/p_b^2\;, \end{align} $$

and thus

$$ \begin{align} \sigma'^2 &=\langle(n'-1)^2\rangle-\langle n'-1\rangle^2 \\ &= (1-p_a)(2-p_b)/p_b^2-(1-p_a)^2/p_b^2\\ &= \frac{1-p_a}{p_b^2}(1+p_a-p_b)\;. \end{align} $$

To check the result, note that it yields the correct values $0$ for $p_a=1$, $(1-p_b)/p_b^2$ for $p_a=0$ and $(1-p_a)p_a$ for $p_b=1$.

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