Q:

Show that if X is a geometric random variable with parameter p, thenE[1/X]= −p log(p)/(1−p)Hint: You will need to evaluate an expression of the formi=1➝[infinity]∑(ai/ i)To do so, writeai/ i=0➝a∫(xi−1) dx then interchange the sum and the integral.

Accepted Solution

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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 $$b=1-p <1$$ we have that:$$\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}$$