We have been discussing pmf. Can someone please show me
A. That the pmf for a binomial random variable sums to 1
B. That the pmf for a geometric random variable sum to 1
So I may see the difference
We have been discussing pmf. Can someone please show me
A. That the pmf for a binomial random variable sums to 1
B. That the pmf for a geometric random variable sum to 1
So I may see the difference
A. You have to show $$\sum_{i=0}^n {n \choose i}p^i(1-p)^{n-i}=1$$ but the left hand side is $$(p+(1-p))^n=1.$$
B. Depending on your definition, you have to show $$\sum_{i=0}^\infty p(1-p)^{i}=1 \text{ or } \sum_{j=1}^\infty p(1-p)^{j-1}=1 $$ and they are clearly the same sums if you replace $i$ by $j-1$.
The geometric series $\sum_{i=0}^\infty x^i = \frac{1}{1-x}$ for $|x|\lt 1$, so (taking the former) $$p\times \sum_{i=0}^\infty (1-p)^{i}=p\times \dfrac{1}{1-(1-p)}=1.$$