8.01 Bernoulli distribution

 

So far we have explored what can be referred to as general discrete random variables. That is, they are not modelled by a particular function or by a particular set of values obtained from a sample space. We will now explore a specific type of discrete random variable - a Bernoulli random variable.

 

Bernoulli random variable

When an experiment can have either of two possible outcomes, usually called success and failure, it gives rise to a Bernoulli random variable. We assign the values $X=1$X=1 and $X=0$X=0 to a Bernoulli random variable $X$X according to whether a trial of the experiment results in a success or a failure. Also, we assign probabilities $p$p and $1-p$1−p to the two outcomes.

Thus, we write $P\left(X=1\right)=p$P(X=1)=p and $P\left(X=0\right)=1-p$P(X=0)=1−p.

 

Is it Bernoulli or is it a general discrete random variable?

Let's consider the experiment of rolling a dice once only.

We recall that when we're exploring a DRV, we choose an interest to focus on. What type of "interest" would constitute a Bernoulli random variable?

If we were to be interested in what appears on the uppermost face of the dice, could this be modelled by a Bernoulli random variable? The answer is no. Why?

When rolling a dice and observing what value lands uppermost on the dice, we have six possible outcomes, not just the two required - a success and a failure. So because the outcomes are NOT only $X=0$X=0 and $X=1$X=1 (and are in fact $X=1$X=1, $X=2$X=2, $X=3$X=3, $X=4$X=4, $X=5$X=5 and $X=6$X=6), this is NOT a Bernoulli random variable.

What if instead we were interested in whether or not a $4$4 lands uppermost on the dice, could this be modelled by a Bernoulli Random Variable? The answer is yes. Why?

This narrows our options down to just two outcomes - either we will see a four, or we will not see a four. We can say that seeing a four land uppermost is a success, and not seeing a four land uppermost is a failure. We would then call the success our outcome to which we assign a $1$1, and the failure our outcome to which we assign a $0$0. So $X=1$X=1 is a four landing uppermost and $X=0$X=0 is anything else landing uppermost.

As with all other discrete random variables, we can represent our distribution using a table or probabilities as shown below.

$x$x	$0$0	$1$1
$P(X=x)$P(X=x)	$\frac{5}{6}$56​	$\frac{1}{6}$16​

 

Practice questions

Question 1

Question 2

 

Expected value, variance and standard deviation of a Bernoulli random variable

The expected value or mean of the Bernoulli random variable $X$X may be thought of informally as the average amount of 'success' per trial over a very large number of trials. This is just $p$p and we write $\mu=p$μ=p or $E(X)=p$E(X)=p. We can confirm this using the formula for the expected value of a discrete distribution and since the outcome for failure is assigned the value of $X=0$X=0, we see it will not influence the value of the mean or expected value.

$E(X)$E(X)	$=$=	$\Sigma xp$Σxp
	$=$=	$0(1-p)+1(p)$0(1−p)+1(p)
	$=$=	$p$p

We can also use our formulas from a general discrete random variable to calculate the variance and standard deviation of a Bernoulli distribution. These values are a measure of the spread of the outcomes of the experiment performed many times. They measure the spread of the distribution in terms of the distance of the outcomes from the mean. For variance, we calculate the expected value of the squared distance of a value from the mean using the formula $Var(X)=E\left((X-\mu)^2\right)$Var(X)=E((X−μ)2).

$Var(X)$Var(X)	$=$=	$E\left((X-\mu)^2\right)$E((X−μ)2)
	$=$=	$(1-\mu)^2p+(0-\mu)^2(1-p)$(1−μ)2p+(0−μ)2(1−p)
	$=$=	$(1-p)^2p+p^2(1-p)$(1−p)2p+p2(1−p)
	$=$=	$p(1-p)(p+1-p)$p(1−p)(p+1−p)
	$=$=	$p(1-p)$p(1−p)

It then follows that the standard deviation of the a Bernoulli random variable is $\sqrt{p(1-p)}$√p(1−p).

 

Practice questions

Question 3

Question 4

Question 5