 # Let m=1/2 and n=1/2 be the probabilities of success and failure respectively and let p(k) be a probability mass function such that p(k)=m \sum_{i=1}^{k}n^i+n \sum_{i=1}^{k} m^i. Celinamg8 2022-09-17 Answered
Let $m=\frac{1}{2}$ and $n=\frac{1}{2}$ be the probabilities of success and failure respectively and let p(k) be a probability mass function such that $p\left(k\right)=m\sum _{i=1}^{k}{n}^{i}+n\sum _{i=1}^{k}{m}^{i}.$
Is this a valid probability mass function?
Say we carry out this test infinitely many times, does $p\left(k\right)=1$?
We recognise that the probability mass function is a sum of two slightly shifted geometric series. Let's consider one of the two (the other will follow the exact same argument).
Say we are summing from 0 to $\mathrm{\infty }$, that is let
${S}_{0}=\sum _{i=0}^{\mathrm{\infty }}{n}^{i}.$
Then this is simply evaluated as ${S}_{0}=1+n+{n}^{2}+{n}^{3}+\cdots =\frac{1}{1-n}$
which is perfectly valid as $n=\frac{1}{2}$. If, then, I start this sequence at $i=1$, is it valid to say that
${S}_{1}=\sum _{i=1}^{\mathrm{\infty }}{n}^{i}=\frac{1}{1-n}-1=\frac{1-\left(1-n\right)}{1-n}=\frac{n}{1-n}$
and as such that, as $k\to \mathrm{\infty }$, we have something like
$\frac{mn}{1-n}+\frac{mn}{1-m}=1.$
Have I made any ridiculous logical jumps or is this valid?
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Step 1
Using your formula we obtain $p\left(1\right)=\frac{1}{2}$ and $p\left(2\right)=\frac{3}{4}$. Thus $p\left(1\right)+p\left(2\right)>1$, which is impossible.
Step 2
If we define the distribution of a random variable X by saying that $Pr\left(X\le k\right)$ is given by your sum, then we will have defined a legitimate distribution. The probability mass function is then given by $p\left(k\right)=m{n}^{k}+n{m}^{k}$. Since $m=n=\frac{1}{2}$, it would be clearer to say $p\left(k\right)=\frac{1}{{2}^{k}}$ for every positive integer k.

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