Hello again!
In this lecture we are going to discuss the Bernoulli distribution.
Before we begin, we use “Bern” to define a Bernoulli distribution, followed by the
probability of our preferred outcome in parenthesis.
Therefore, we read the following statement as “Variable “X” follows a Bernoulli
distribution with a probability of success equal to “p””.
Okay!
We need to describe what types of events follow a Bernoulli distribution.
Any event where we have only 1 trial and two possible outcomes follows such a distribution.
These may include a coin flip, a single True or False quiz question, or deciding whether
to vote for the Democratic or Republican parties in the US elections.
Usually, when dealing with a Bernoulli Distribution, we either have the probabilities of either
event occurring, or have past data indicating some experimental probability.
In either case, the graph of a Bernoulli distribution is simple.
It consists of 2 bars, one for each of the possible outcomes.
One bar would rise up to its associated probability of “p”, and the other one would only reach
“1 minus p”.
For Bernoulli Distributions we often have to assign which outcome is 0, and which outcome
is 1.
After doing so, we can calculate the expected value.
Have in mind that depending on how we assign the 0 and the 1, our expected value will be
equal to either “p” or “1 minus p”.
We usually denote the higher probability with “p”, and the lower one with “1 minus
p”.
Furthermore, conventionally we also assign a value of 1 to the event with probability
equal to “p”.
That way, the expected value expresses the likelihood of the favoured event.
Since we only have 1 trial and a favoured event, we expect that outcome to occur.
By plugging in “p” and “1 minus p” into the variance formula, we get that the
variance of Bernoulli events would always equal “p, times 1 minus p”.
That is true, regardless of what the expected value is.
Here’s the first instance where we observe how elegant the characteristics of some distributions
are.
Once again, we can calculate the variance and standard deviation using the formulas
we defined earlier, but they bring us little value.
For example, consider flipping an unfair coin.
This coin is called “unfair” because its weight is spread disproportionately, and it
gets tails 60% of the time.
We assign the outcome of tails to be 1, and p to equal 0.6.
Therefore, the expected value would be “p”, or 0.6.
If we plug in this result into the variance formula, we would get a variance of 0.6, times
0.4, or 0.24.
Great job, everybody!