Finding the Probability of a random variable with countably infinite values

[swastik]

Here is the question I am working on:
Let X be a discrete random variable with probability mass function pX(k) = c/3^k for k = 1, 2, ... for some c > 0.
Find c.
Find P(X ≥ k) for all k = 1, 2,3.....

Now the first part is easy and the value of c comes out to be 2.
How do I solve the second part? Any suggestions?

 

[HallsofIvy]

Yes, you are given that $\displaystyle P(1)= \frac{c}{3}$, $\displaystyle P(2)= \frac{c}{3^2}=\frac{c}{9}$, etc. The condition on c is that the "total" probability is $\displaystyle \frac{c}{3}+ \frac{c}{9}+ \cdot\cdot\cdot+ \frac{c}{3^n}+ \cdot\cdot\cdot= c(\frac{1}{3}+ \frac{1}{3^3}+ \cdot\cdot\cdot+ \frac{1}{3^n}+ \cdot\cdot\cdot)$. That is a geometric series except that it begins at n= 1 rather than n= 0. The sum of a geometric sequence with initial term c and "common ratio" $\displaystyle \frac{1}{3}$ is $\displaystyle \frac{c}{1-\frac{1}{3}}= \frac{3c}{2}$. Since this series is missing the first term, c, its sum is $\displaystyle \frac{3c}{2}- c= \frac{c}{2}$. That must be 1 so, yes, $\displaystyle c= 2$.

Now, the probability that $\displaystyle X\ge k$ is simply that minus the probability that $\displaystyle X< k$. The probability that $\displaystyle X< k$ is the sum $\displaystyle \frac{2}{3}+ \frac{2}{9}+ \cdot\cdot\cdot+ \frac{2}{3^{k-1}}$ which is a finite geometric sum. The sum of such a finite geometric sum, with initial term a and common ratio r, from 0 to n, is $\displaystyle \frac{a(1- r^{n+1})}{1- r}$. Here, a= 2 and r= 1/3 with n= k- 1 so that is $\displaystyle \frac{2(1- (\frac{1}{3})^{k})}{1-\frac{1}{3}}= (3/2)(1- (\frac{1}{3})^{k})$. Subtract that from 1 to get $\displaystyle P(X\ge k)$.

 

[swastik]

Thanks a lot. That cleared my doubts