How Does Expected Value Work?

The Expected Value

>Mean and Expected Value>How Does Expected Value Work?

Theory

$P (X = x_{i})$

$\frac{1}{6}$

$\frac{1}{6}$

$\frac{1}{6}$

$\frac{1}{6}$

$\frac{1}{6}$

$\frac{1}{6}$

The expected value is, in the long run, the same as the mean,. It is calculated in this way:

μ = E (X) = \sum_{i = 1}^{m} x_{i} \cdot P (X = x_{i})

Example 1

You throw a die. What is the expected value of the number you roll?

Make a table with the possible outcomes and their probabilities:

\begin{array}{l} μ = E (X) & = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} \\ + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} \\ = 3.5 \end{array}

μ = E (X) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + 3 \cdot \frac{1}{6} + 4 \cdot \frac{1}{6} + 5 \cdot \frac{1}{6} + 6 \cdot \frac{1}{6} = 3.5

The expected value is

E (x) = 3.5

Example 2

You insure your laptop with an insurance company. The policy of the insurance company is to only compensate you under two circumstances:

Your laptop is stolen.

Your laptop is broken.
In the first case you’ll be compensated $$ 400$ , while in the second case you’ll be compensated $$ 200$ .
Assume the probability of your laptop being stolen is $2 %$ , and the probability of your laptop breaking is $4 %$ . To keep it simple, let’s assume that these probabilities are the same any given year.

1.
Let $X$ be the compensation you receive from the insurance company in a year. Find the expected value of $X$ .
2.
What is the least premium you would have to pay in order for the insurance company to not lose money on this insurance policy?

1.

From the assignment you have

P (X = laptop stolen) = 0.02

and

P (X = laptop broken) = 0.04

The expected compensation during any given year then becomes

\begin{array}{l} E (X) & = 400 \cdot 0.02 + 200 \cdot 0.04 \\ = $ 16 \end{array}

E (X) = 400 \cdot 0.02 + 200 \cdot 0.04 = 16 .

2.: As the expected compensation any given year is $ $16$ , the premium has to be at least $ $16$ a year for the insurance company to not lose money.

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