# How to prove that $f_X(x) = (1-p)^xp$ is a probability function?

## Exercise:

An experiment is performed where an event A (success) with probability p, or no occurrence (failure) can occur, with probability 1 - p, where p ∈ (0, 1). In successive repetitions of the same experiment it will be assumed that the probability p is kept constant in each of them and, in addition, that they are independent.

Consider the discrete random variable X, defined as the "number of failures before obtaining the first success in successive experiments".

a) What is the range of this random variable?

b) Deduce the analytic expression of the probability function of this random variable.

c) Show that the function obtained in b, is indeed a probability function.

Note: Any discrete random variable, with the above probability function, is called the Geometric random variable, of parameter p, and is represented by X ~ G (p).

## Solution:

• a) $R_X = {0,1,2,3,...}$
• b) $f_X(x) = (1-p)^xp$
• c) ?
• You need to show that $$\sum_{k=0}^\infty (1-p)^kp = 1$$ – Theoretical Economist Feb 5 '17 at 23:05
• @Thomas Yes, that was a typo. Whoops. Fixed now. – Theoretical Economist Feb 5 '17 at 23:07

$f_X$ is nonnegative and its mass is $\sum_{x=0}^\infty f_X(x)=p\frac{1}{1-(1-p)}=1$ (I've used the geometric series in the first equality) so $f_X$ is indeed a probability function.
• You said "so $f_X$ is indeed a probability function". Does that mean you've proven it? – emi Feb 5 '17 at 23:41
• @CarlosFrostte Yes, in order to check that $f_X$ is a probability function you need to verify two things: that $f_X(x)\geq0$ for every $x$ (which I only asserted at the beginning since in this case it is trivial), and that its mass (i.e. the sum of $f_X(x)$ over $x$) is $1$ (which I calculated in the answer). – user378947 Feb 5 '17 at 23:46
1) $P(X=x)\ge 0$. $\to (1-p)^xp>0$ for all $x = 0,1,2,3...$
2) $P(\Omega)=1$. $\to P(\Omega) = \sum_{x=0}^{\infty}(1-p)^xp = \frac{p}{1-(1-p)}=1$.