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Iteru [2.4K]
3 years ago
5

Show that if X is a geometric random variable with parameter p, then

Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

\sum_{k=1}^{\infty} b^{k-1}

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

And with this we have the requiered proof.

And since b=1-p we have that:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

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