# Another probability distribution question:

So apparently I'm having quite a difficult time finding probability distributions. My first question is how to develop a strategy of finding these probability distributions. My next question comes from a book I'm using to study for the CAS Exam P and it goes like this:

Let $p(x)=(\frac34)(\frac14)^x, x=0,1,2,3...$ be probability mass function of a random variable $X$. Find $F$, the distribution of $X$.

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Who downvoted this? I mean, I can guess it was glebovg, but since we're dealing with probability, I would have to say I'm only 95% sure of it. There's no reason to downvote this. If you have a reason, why not explain it to the OP so the OP knows what he can do to improve? –  Graphth Oct 25 '12 at 20:27
Kyle, did any of the answers given answer your question? –  Graphth Oct 28 '12 at 16:11
Yes thank you @Graphth and Hagen I would accept both of your answers as the correct answer if I could. –  TheHopefulActuary Oct 28 '12 at 22:31
This is a geometrique distribution. –  user31280 Oct 29 '12 at 0:35
If you do not know how to find cdf by now, you should not become an actuary. –  glebovg Oct 29 '12 at 18:25

## 2 Answers

Hagen answered a specific part of your question, but I see that your question is how to understand how to do this in general. So, that is why I am answering the question. Please let me know if there are parts that are not clear.

In general, the cumulative distribution function is defined by

$$F(x) = P(X \leq x).$$

You already have $p(x) = P(X = x)$. So, if you want to find $F(n) = P(X \leq n)$, you just add up all of the values of $p(x)$ for all $x \leq n$. That is

\begin{align*} F(n) &= P(X \leq n) \\ &= P(X = 0) + P(X = 1) + P(X = 2) + \cdots + P(X = n) \\ &= p(0) + p(1) + p(2) + \cdots + p(n). \end{align*}.

The above formula assumes that $n$ is an integer. But, if $n$ were not an integer, you would just add up all values of $p(x)$ for $x \leq n$, just as I said. For example,

$$F(3.14159) = p(0) + p(1) + p(2) + p(3).$$

Now, if you want to find a nice formula for $F(x)$ for this specific problem, see Hagen's answer. In general, nice formulas may or may not be possible, and the techniques for finding them would vary hugely from problem to problem.

For this specific problem, note, for any $x$ in $[0, 1)$, $F(x)$ will have the same value, $p(0)$. And, for any $x$ in $[1, 2)$, $F(x) = p(0) + p(1)$. And, for any $x$ in $[2, 3)$, $F(x) = p(0) + p(1) + p(2)$. And so on. So, $F(x)$, as a function, has an infinite number of jump discontinuities. The main thing here is, for any $x$ value, hopefully you now understand how to find $F(x)$.

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glebovg, when you downvote my answer, please explain why you are doing it. –  Graphth Oct 26 '12 at 14:55
Precisely what I explained in my comment below. –  glebovg Oct 26 '12 at 18:47
@glebovg Actually, that's not even close to true. This barely goes into the specifics of the specific problem he asked about. It answers the general question he asked about. My answer gives the definition of a cdf and then talks about what a general cdf is like for a discrete distribution. Essentially, you would downvote my answer no matter what. And, you know it. And, I knew it too. –  Graphth Oct 26 '12 at 21:05
@glebovg The point being, I obviously have not done anyone's homework here. Even if it were tagged homework and he just asked for a hint, this would be general enough as to be acceptable. I barely gave more than the definition. But, it is clear this is not a homework question, so that shouldn't matter. –  Graphth Oct 26 '12 at 21:15

$F(x)$ is the probability of $X\le x$. Thus $F(x)=0$ for $x<0$, $F(x)=\frac34$ for $0\le x<1$, $F(x)=\frac{15}{16}$ for $1\le x<2$. You should find that $$F(x)=1-\frac1{4^{\lfloor x\rfloor+1}}$$ if $x\ge0$.

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Why the downvote? –  Hagen von Eitzen Oct 25 '12 at 6:11
You are supposed to help the OP understand the concept, but you simply did all the work. –  glebovg Oct 25 '12 at 15:16
@HagenvonEitzen glebovg loves to downvote, around 80% of his votes. He believes his answer that has absolutely nothing to do with the question is more useful than your answer which actually answers the question. –  Graphth Oct 25 '12 at 20:24
If you did not know voting is a feature of StackExchange. I believe my answer has something to do with the question whereas your comments do not. –  glebovg Oct 25 '12 at 21:05
@glebovg Your answer has nothing to do with the question. –  Graphth Oct 26 '12 at 14:32