# Sample space of pmf

I have several exercises to solve that deal with sample space of PMF.

One is:

Let $X_1,\dots , X_n$ be independent random variables with pmf $p(x;\pi) = (1-\pi)^x \pi$

What is the sample space of $X_1?$ Try to give a probabalistic interpretation of such a sample space. Hint: for example. a Bernoulli random variable can be used to model a coin with probability of success $p \in \;\rbrack0,1\lbrack$.

Any hint? I do not know how to "see" from the distribution what the sample space is without knowing more about this distribution...

## Possible Solution?

I have the following idea: If I plug in all values in the sample space and calculate the sum I have to get $1$ (because it is a probability). But obviously here we can have just one value for the sample space which is $\log({\frac{1}{\pi}})/{\log(1-\pi)}$ (I set the formula for pmf to one and solve the equation for x).

Is that correct?

• This question is not answerable because the domain of the pmf has not been specified. (I suspect there is a typographical error and that the pmf might be $(1-\pi)^{1-x}\pi^x$ for $x\in\{0,1\}$.)
– whuber
Oct 15 '13 at 17:47
• The exercise IS realy like this... I was also puzzled by this, but I can not say why it looks odd. There is no solution for sure? Oct 15 '13 at 17:57
• A lot of exercises like this are really more designed (in my experience) to emphasize or test recognition of the pmf. So we'd assume that $p(x;\pi)$ is some slight specialization of a standard pmf, perhaps by a simple transform of the random variable, try to figure out which one it is, then get the sample space from the definition of the (standard) pmf. In this case, the random variable $x$ can be easily transformed to another random variable, let us say $y$, for which $p(y;\pi)$ is a well-known distribution, and everything else follows from that. Oct 15 '13 at 18:07

There are two huge problems with this question that make it unanswerable.

First, a sample space is a set of outcomes of an experiment. A random variable is a function assigning a unique real value to each outcome. A probability measure on the sample space determines the distribution of the random variable. Given only the distribution, we cannot possibly identify the sample space. For instance, the random variable assigning the value $0$ to "tails" and $1$ to "heads" to describe outcomes of the flip of a fair coin has sample space {"heads", "tails"}. It has a Bernoulli$(1/2)$ distribution. The random variable assigning the value $0$ to all points in the Earth's northern hemisphere and $1$ to all points in its southern hemisphere has a sample space consisting of all points on Earth. If all points are considered equally probable, then this variable, too, has a Bernoulli$(1/2)$ distribution--but obviously points on the Earth are not flips of a coin!

Second, let's re-interpret the question to ask about the set of possible values of a random variable (its range as a function), because maybe there is a chance this could be answered with the information given. Unfortunately, the question is still ambiguous. This can be shown by exhibiting two different distributions with different ranges that nevertheless satisfy the given conditions: namely, there exists some number $\pi$ such that the probability of each possible value $x$ is given by the formula $(1-\pi)^x\pi.$ Such a formula suggests (but does not explicitly indicate) that $x$ is intended to be integral, which helps limit our search for counterexamples.

1. Let the range be $\{-1, 0\}$. The total probability is

$$1 = (1-\pi)^{-1}\pi + (1-\pi)^0\pi.$$

One solution is $\pi = (3 - \sqrt{5})/2 \approx 0.381966.$

2. Let the range be $\{-1, 0, 1\}$. The total probability is

$$1 = (1-\pi)^{-1}\pi + (1-\pi)^0\pi + (1-\pi)^1\pi.$$

One solution is a root of $x^3 - 3 x^2 + 4 x - 1$ approximately equal to $0.317672.$ (It is the only root lying between $0$ and $1.$)

The first distribution assigns probabilities $0.618034$ to $-1$ and $0.381966$ to $0$; the second distribution assigns probabilities $0.465571$ to $-1$, $0.317672$ to $0$, and $0.216757$ to $1$: obviously they are different and have different ranges.

• I like the way your answer skirts around the answer that most people would arrive at. Oct 16 '13 at 2:22
• @Dilip I have no clue what that might mean. What answer to you believe most people would offer?
– whuber
Oct 16 '13 at 2:52
• I would have said that $X$ is a geometric random variable with parameter $\pi$, where $\pi$ does not have the usual meaning that it has in mathematical circles, pun intended, but is a number in $(0,1)$, and $x$ takes on all nonnegative integer values. Using the hint, the experiment would consist of tossing a biased coin with $P(\text{Heads})=\pi$ until a Head occurs, the sample space would be the set $$\Omega=\{H,TH,TTH,\ldots\}$$ with outcomes having probabilities $\pi,(1-\pi)\pi,\ldots$, and $X$ being the number of Tails in the outcome. Oct 16 '13 at 3:07
• @Dilip I see, thank you. It is of note that you had to supply several assumptions that, although they may be obvious to a trained statistician, are certainly not explicit in the question. I have instead attempted to take this question at face value--much as a neophyte would--and answer it on its merits (which are few indeed :-).
– whuber
Oct 16 '13 at 16:46
• Alas, I am not a trained statistician, and yes, I had to supply several assumptions based on the Hint: for example. a Bernoulli random variable can be used to model a coin with probability of success $p \in ]0,1[$, that is in the problem statement. To me, the natural extension was to a geometric random variable. I agree with you about the merits of the question; indeed, even the notation is quite dreadful. $\pi_0$ and $\pi_1$ as the prior probabilities of two hypotheses is common usage, but just $\pi$? with all the confusion likely to ensue when normal distributions are encountered? Oct 16 '13 at 18:30