I am having trouble understanding the intuition behind the CDF and survival probability of a geometric distribution on both $\{0, 1, \ldots \}$ and on $\{1, 2, 3, \ldots \}$.

I know that a geometric starting from $0$ is the number of failures before the first success and the geometric starting at 1 is the number of failures including the first success.

Can someone please explain the intuition behind the CDF and survival functions and explain what the formula is?

I think that the part confusing me is: Why is the following not true? $$p\left(X>n\right)=\left(1-p\right)^{n+1}+\left(1-p\right)^{n+2}+...$$ for a geometric starting at $0$.

  • $\begingroup$ I guess it would be more helpful if you consult your textbook first and come then with specific questions about the CDF and survival probabilities where you have difficulties. Nobody will be able ( and probably not willing) to explain you here these concepts in a short mail. $\endgroup$ – Karl Oct 28 '18 at 15:18
  • $\begingroup$ One has to wonder what that equation means. For starters, you have an $x$ on the LHS while the RHS is in terms of $p$ and $n$. Also, please capitalize your measure too as $P(X>n)$. Notation is important and makes life much easier and clearer, I swear. $\endgroup$ – Nap D. Lover Oct 28 '18 at 16:32

If $X$ is the number of failures before the arrival of first success then the event $\{X>n\}$ is the same as the event that the first $n+1$ trials are not a success, so the probability on that is $(1-p)^{n+1}$.

Then consequently:$$F_X(n)=P(X\leq n)=1-P(X>n)=1-(1-p)^{n+1}$$

Similarly we find: $$F_X(n)=P(X\leq n)=1-P(X>n)=1-(1-p)^{n}$$if $X$ denotes the number of trials need to arrive at the first success.

Concerning your question note that $(1-p)^{n+k}$ is the probability that the first $n+k$ trials are failures.

However these events for $k=1,2,3,\dots$ are not mutually exclusive.

So summation of the terms leads to multiple counting.

| cite | improve this answer | |
  • $\begingroup$ Why isn't P[X > n] = (1-p) ^n+1 + (1-p)^n+2 +.... $\endgroup$ – user485656 Oct 28 '18 at 15:25
  • $\begingroup$ I have edited my answer in order to explain that. $\endgroup$ – drhab Oct 28 '18 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.