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?
 A: 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.


*

*Let the range be $\{-1, 0\}$.  The total probability is
$$1 = (1-\pi)^{-1}\pi + (1-\pi)^0\pi.$$
One solution is $\pi = (3 - \sqrt{5})/2 \approx 0.381966.$

*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.
