# Intuition behind P(X>x) for a Geometric distribution

It can be formally proved that for a Geometric distribution with parameter p and pmf $$P_X(x)=(1-p)^{x-1} p$$, then $$P(X>x) = (1-p)^x$$.

One counts the failures up to $$X=x$$ and that's it. Intuitively I don't understand why it's right.

Because for $$(X>x)$$ one has the sum of infinite options. I mean that after the $$(1-p)^x$$ failures one can have the succes.

Or after $$(1-p)^{x+1}$$ then you can have the success or after $$(1-p)^{x+2}$$ and so on.

I think that the probability of $$(X>x)$$ would be the sum of $$p[(1-p)^x + (1-p)^{x+1} + (1-p)^{x+2} + (1-p)^{x+3}$$ ... up to infinity$$]$$. From an intuitively point of view, what's wrong with this argument?

Your argument is essentially $$P(X>x) = \sum_{k=x+1}^\infty P(X=k)$$ which is correct.

To explicitly verify, we can check that your last expression is equal to the desired result. $$p \sum_{k=x}^\infty (1-p)^k = p(1-p)^x \sum_{m=0}^\infty (1-p)^m = \frac{p(1-p)^x}{1-(1-p)} = (1-p)^x.$$

• Thanks.I'm trying to understand it. If my last expression is $\sum_{k=x+1}^\infty P(X=k)$. Why in the next step you start the sum $p \sum_{k=x}^\infty (1-p)^k$ with $k=x$ ? Wouldn't it be $k=x+1$? – roy212 May 2 at 4:55
• @roy212 The last expression you wrote at the end of your post is correct so you can just start there. (But if you are still curious, you can write $P(X > x) = \sum_{k=x}^\infty P(X = k+1)$ instead...) – angryavian May 2 at 4:58

One counts the failures up to X=x and that's it. Intuitively I don't understand why it's right.

A geometric random variables represents a count of iid Bernoulli trials until the first success.   The event of $$\{X>x\}$$ is that this count exceeds $$x$$, which is the event of having at least $$x$$ consecutive failures prior to the first success.

Thus $$\mathsf P(X>x)=(1-p)^x$$.

I think that the probability of (X>x) would be the sum of $$p[(1−p)^x+(1−p)^{x+1}+(1−p)^{x+2}+(1−p)^{x+3}$$ ... up to infinity]. From an intuitively point of view, what's wrong with this argument?

Nothing.

Now as you say, we should anticipate that this equals $$\sum_{k=1}^\infty \mathsf P(X=x+k)$$.

$$\mathsf P(X>x){=\sum_{k=1}^\infty\mathsf P(X=x+k)\\=\sum_{k=1}^\infty (1-p)^{x+k-1}p\\=(1-p)^x p\sum_{k=1}^\infty(1-p)^{k-1}\\=(1-p)^xp\sum_{k=0}^\infty(1-p)^{k}}$$

Okay, so let $$S=\sum_{k=0}^\infty(1-p)^{k}$$, which will converge to a real value when $$\lvert (1-p)\rvert \leq 1$$. This famous series is called the Geometric Series. ... this is from where a Geometric Distribution obtains its name.

$$S={1+\sum_{k=1}^\infty (1-p)^k\\=1+(1-p)\sum_{k=0}^\infty(1-p)^k\\=1+(1-p)S}$$ Which means that $$S=1/p$$ and hence $$\mathsf P(X>x)=(1-p)^x$$

• "The event of $\{X>x\}$ is that this count exceeds $x$, which is the event of having at least $x$ consecutive failures prior to the first success.Thus $\mathsf P(X>x)=(1-p)^x$" When I think of "at least three" I automatically think that there can be three or more. I think of "at least three" the way you stated when the event is "exactly three". In my opinion, this argument is not very convincing. Just my thoughts. – roy212 May 2 at 8:42
• $\{X>x\}$ is the event of counting more than $x$ trials until the first success, or that the count of failures before the first success is at least $x$. Thus it is the event that the first $x$ trials are all failures (with the first success occurring somewhere later). The probability that the first $x$ trials are all failures is.… @roy212 – Graham Kemp May 2 at 9:03
• If the count of trials until the first success exceeds three, then the first three trials must be failures (there may be more, but at least those will be). If the first three trials are failures, then the count of trials until the first success must exceed three. They are the same event, and denoted as $\{X>3\}$. – Graham Kemp May 2 at 9:11