The Geometric Distribution (Edexcel A Level Further Maths: Further Statistics 1): Revision Note

Conditions for geometric models

What is the geometric distribution?

• The geometric distribution models the number of trials needed to reach the first success

  • For example, how many times will you have to roll a dice until it lands on a '6' for the first time

• The notation for the geometric distribution is $\mathrm{Geo}\left(p\right)$ 

  • For a random variable $X$ that has the geometric distribution you can write $X~\mathrm{Geo}\left(p\right)$

  • $X$ is the number of trials it takes to reach the first success

    • For example, $X=8$ means the first success occurred on the 8th trial

  • $p$ is the fixed probability of success in any one trial

What are the conditions for using a geometric model?

• A geometric model can be used for an experiment that satisfies the following conditions:

  • The experiment consists of an indefinite number of successive trials

  • The outcome of each trial is independent of the outcomes of all other trials

  • There are exactly two possible outcomes for each trial (success and failure)

  • The probability of success in any one trial ($p$) is constant 

•  Note that these conditions are very similar to the conditions for the binomial distribution

  • But for a binomial distribution the number of trials ($\mathit{n}$) is fixed

    • And you count the number of successes

  • While for a negative binomial distribution the experiment continues until the first success is achieved

    • And you count the number of trials it takes to reach that first success

When might the conditions not be satisfied?

• If asked to criticise a geometric model, you may be able to question whether the trials are really independent

  • For example, someone may be repeating an activity until they achieve a success

    • The trials may not be independent because the person gets better from practising the activity

    • This also means the probability of success, $p$, is not constant

  • In order to proceed using the model, you would have to assume that the trials are independent

Geometric probabilities

What are the probabilities for the geometric distribution?

• If $X~\mathrm{Geo}\left(p\right)$ then $X$ has the probability function:

  • $\mathrm{P}\left(X=x\right)=\mathrm{p}\left(x\right)\mathit{=}p{\mathit{(}1\mathit{-}p\mathit{)}}^{x-1}\mathit{,}x=1\mathit{,}2,3,...$

  • the random variable $X$ is the number of trials needed to get the first success

  • $p$ is the constant probability of success in one trial

  • $\mathrm{P}\left(X=x\right)$ is the probability that the first success will occur on the $x\mathrm{th}$ trial

• Note that that is the product of

  • the probability of first getting $\mathit{x}\mathbf{-}\mathbf{1}$ failures,  ${\mathit{(}\mathit{1}\mathit{-}p\mathit{)}}^{x\mathit{-}1}$,

  • and the probability of getting a success in the $\mathit{x}\mathbf{th}$ trial, $p$

• Also note that there is no greatest possible $x$

  • It could require any number of trials to reach the first success

  • However $\mathrm{P}\left(X=x\right)$ gets closer and closer to zero as $x$ gets larger

• Your calculator may allow you to calculate Geometric probabilities directly

  • i.e., without having to use the above formula

What are the properties of the geometric distribution?

• Note that $\mathrm{P}\left(X=1\right),\mathrm{P}\left(X=2\right),\mathrm{P}\left(X=3\right),\mathrm{P}\left(X=4\right),...=p,p\left(1-p\right),p{\left(1-p\right)}^2,p{\left(1-p\right)}^3,...$

  • This means that the geometric probabilities form a geometric sequence

    • The first term is $p$

    • The common ratio is $\left(1-p\right)$

  • This is where the geometric distribution gets its name!

• Assuming that $0<p<1$, then it is also true that $0<\left(1-p\right)<1$

  • This means that  $\mathrm{P}\left(X=1\right)>\mathrm{P}\left(X=2\right)>\mathrm{P}\left(X=3\right)>\mathrm{P}\left(X=4\right)>...$

    • i.e., the probabilities form a decreasing sequence

    • and $\mathrm{P}\left(X=1\right)$ is the largest probability in the sequence

    • Therefore $X=1$ is the mode of the distribution

• The geometric distribution has no 'memory'

  • It doesn't matter what has happened previously, or how many 'failures' in a row there have been

  • The probability of getting a 'success' in any trial is always $p$

  • This means that the number of additional trials needed for the first success is not dependent on the number of trials that have already occurred

    • e.g. if 5 (failed) trials have already occurred, the probability of the first success happening after 7 trials is simply the probability of success happening after 2 trials in the first place, i.e. $p\left(1-p\right)$

What are the cumulative probabilities for the geometric distribution?

• If $X~\mathrm{Geo}\left(p\right)$ then $X$ has the cumulative geometric distribution:

  • $\mathrm{P}\left(X\le x\right)=1-{\mathit{(}1\mathit{-}p\mathit{)}}^x\mathit{,}x=1\mathit{,}2,3,...$

  • the random variable $X$ is the number of trials needed to get the first success

  • $p$ is the constant probability of success in one trial

  • $\mathrm{P}\left(X\le x\right)$ is the probability that the first success will occur on or before the $x\mathrm{th}$ trial

• Your calculator may allow you to calculate Geometric probabilities directly

  • i.e., without having to use the above formula

• The formula can be proved as follows

  • If the first success occurs on or before the $x\mathrm{th}$ trial, that means that the first $x$ trials have not all been failures

    • The probability of getting $x$ failures in a row is ${\mathit{(}1\mathit{-}p\mathit{)}}^x$

    • So the probability of that not happening is $1-{\left(1-p\right)}^x$

• Alternatively, it can be proved algebraically

  • The geometric probabilities form a geometric sequence with first term $p$ and common ratio $1-p$

  • Putting that into the geometric series formula $S_n=\frac{a\left(1-r^n\right)}{1-r}$ gives

$\mathrm{P}\left(X\le x\right)=\sum _{r=1}^{x}P\left(X=r\right)=\frac{p\mathit{(}\mathit{1}\mathit{-}{\mathit{(}\mathit{1}\mathit{-}p\mathit{)}}^x\mathit{)}}{1-\left(1-p\right)}=\frac{p\mathit{(}\mathit{1}\mathit{-}{\mathit{(}\mathit{1}\mathit{-}p\mathit{)}}^x\mathit{)}}{p}=1-{\left(1-p\right)}^x$

• Because $\underset{x\to \infty }{\lim }{\left(1-p\right)}^x=0$ (assuming $0<p<1$), it follows that

$\sum _{r=1}^{\infty }\mathrm{P}\left(X=r\right)=1-0=1$

• So the sum of all probabilities is equal to 1

  • This is a requirement of any probability distribution

Geometric mean & variance

What are the mean and variance of the geometric distribution?

• If $X~\mathrm{Geo}\left(p\right)$, then

  • The mean of $X$ is  $\mathrm{E}\left(X\right)=\mu \mathit{=}\frac{1}{p}$

• The variance of $X$ is  $\mathrm{Var}\left(X\right)={\sigma }^{\mathit{2}}\mathit{=}\frac{\mathit{1}\mathit{-}\mathit{p}}{p^{\mathit{2}}}$

• You need to be able to use these formulae to answer questions about the geometric distribution