Explain and calculate expected value and higher moments, mode, median, and percentile

Expected Value of Discrete Random Variables

Let \(X\) be a discrete random variable with probability mass function, \(p(x)\).

The expected value or the mean of the random variable \(X\), denoted as \(E(X)\), is given by:

$$ E \left(X\right)=\sum{x. p (x)} $$

Where the summation is taken over all possible values of \(X\).

The expected value of \(X\) can also be denoted as \(\mu\) or \(\mu X\).

The expected value of random variable \(X\) is often written as \(E(X)\) or \(\mu\) or \(\mu X\)

Now, let us discuss some basic properties of mathematical expectations.

Example: Expected Value of Discrete Random Variables

A dice is rolled, and you win the number of dollars shown on the dice. Find the expected value of your winnings.

Solution

To find the expected value of your winnings, you’ll need to multiply each possible outcome by its probability and then sum up those products:

$$ \begin{array}{c|c|c|c|c|c|c}
\text{Outcome (X)} & \$1 & \$2 & \$3 & \$4 & \$5 & \$6 \\ \hline
\text{Probability (P(X))} & \frac {1}{6} & \frac {1}{6} & \frac {1}{6} & \frac {1}{6} & \frac {1}{6} & \frac {1}{6} \end{array} $$

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

So, the expected value of your winnings when you roll a fair six-sided dice is $3.50.

Example 2: Expected Value of Discrete Random Variables

A short-term disability insurance policy will pay a benefit for each week of disability. The benefit is $250 per week, up to 4 weeks, and $150 per week thereafter. The number of weeks of disability, \(Y\), is a discrete random variable with:

$$ p(y)= \left\{\begin{matrix} \frac{1}{55}(6-y)^2 & \text{for } y=1,2,3,..6 \\0 & \text{elsewhere} \end{matrix}\right. $$

Calculate the expected value of the total benefit payment.

Solution

The simplest approach is to create a table to illustrate the benefits and corresponding probabilities:

$$ \begin{array}{c|c|c|c|c|c|c}
\text{Weeks (y)} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline
\text{Benefit} & \$250 & \$500 & \$750 & \$1,000 & \$1150 & \$1,300 \\ \hline
P(y) & {{=\dfrac {1}{55}\times(6-1)^2 }\\ {=\dfrac {25}{55}} } & \dfrac {16}{55} & \dfrac {9}{55} & \dfrac {4}{55} & \dfrac {1}{55} & 0
\end{array} $$

We then compute the mean as:

$$ \begin{align*} E(X) & = 250 \times \frac {25}{55}+500 \times \frac {16}{55}+750 \times \frac {9}{55} \\ & +1,000 \times \frac {4}{55}+1,150 \times \frac {1}{55} \\ & =\$475.45 \end{align*} $$

Alternatively, we can compute the mean as follows:

$$ \begin{align*}
E(X) & =\sum_{y=1}^4 250y\left(\frac {1}{55} \right) (6-y)^2+\sum_{y=5}^6 (600+150y)\left(\frac {1}{55} \right) (6-y)^2 \\ & =250 \times 1 \times \left(\frac {1}{55} \right) \times 25+250 \times 2 \times \left(\frac {1}{55} \right) \times 16 \\ & +250 \times 3 \times \left(\frac {1}{55} \right) \times 9+250 \times 4 \times \left(\frac {1}{55} \right) \\ & \times 4+(400+150(5))\left(\frac {1}{55} \right)+0 \\
& =113.64+145.45+122.73+72.72+20.91 \\ & = $475.45 \end{align*} $$

Therefore, the expected benefit payment from the short-term disability insurance policy is approx. $475.

Example 3:

Properties of Mathematical Expectations

If \(\alpha\) and \(\beta\) are constant, then:

1. \(E(\alpha)=a\); i.e., the expected value of a constant \((\alpha)\) is simply the constant itself.
2. \(E(\alpha X)=a.E(X)\)
3. \(E(\alpha X+\beta)=\alpha E(X)+\beta\)

These properties are particularly important when we are dealing with constants, or random variables that have constants.

Example: Expected Return of a Discrete Random Variable (Rolling a Fair Die):

Suppose we are conducting an experiment of rolling a fair six-sided die. Let \(X\) be the random variable representing the outcome of rolling the die.

1. Find \(E(X)\).
2. Find \(E(3X+10)\).

Solution

1. $$ \begin{array}{c|c}
   X & P(X=x) \\ \hline
   1 & \frac {1}{6} \\ \hline
   2 & \frac {1}{6} \\ \hline
   3 & \frac {1}{6} \\ \hline
   4 & \frac {1}{6} \\ \hline
   5 & \frac {1}{6} \\ \hline
   6 & \frac {1}{6}
   \end{array} $$

   We know that,

   $$ E(X)=\sum x.p (x) $$

   In this case,

   $$ \begin{align*} p(x)& =P(X=x)= \left\{\begin{matrix} \frac {1}{6}, & x=1,2,3,4,5,6 \\0, & \text{Elsewhere} \end{matrix} \right. \\
   \therefore E(X) & = 1 \times \frac {1}{6}+2 \times \frac {1}{6}+3 \times \frac {1}{6}+4 \times \frac {1}{6}+5 \times \frac {1}{6}+6 \times \frac {1}{6}= 3.5 \end{align*} $$

2. Using the properties of expectation,

   $$ E(3X+10)=3E(X)+E(10)=3 \times 3.5+10=20.5 $$

   Alternatively,

   $$ \begin{align*} E(3x+10) & =(3\times 1+10) \times \frac {1}{6}+(3 \times 2+10) \times \frac {1}{6} \\ & +(3 \times 3+10)\times \frac {1}{6}+(3 \times 4+10)\times \frac {1}{6} \\ & +(3\times 5+10) \times \frac {1}{6}+(3 \times 6+10) \times \frac {1}{6} \\ & =20.5 \end{align*} $$

This property shows that the expected value is a linear operator.

The following are commonly tested discrete distributions:

• Poisson Distribution
• Binomial Distribution (Special Case: Bernoulli Distribution)
• Negative Binomial Distribution (Special Case: Geometric Distribution)
• Discrete Uniform Distribution
• Hypergeometric Distribution
• Multinomial Distribution

In this reading, we will give a summary of the probability mass functions of the above-mentioned distributions as well as the formula for their expectations. In the next reading, however, we will discuss each distribution in more detail.

$$ \small{\begin{array}{l|c|c|l}
\textbf{Distribution} & \textbf{PMF} & \textbf{Expectation} & \textbf{Parameters} \\ \hline
\text{Poisson} & P(X=x)=\frac {\lambda^x e^{-\lambda}}{xǃ},x=0,1,2,,\dots & E(X)=\lambda &
\lambda= \text{Rate parameter} \\ \hline
\text{Binomial} & p(X=x)=\binom{n}{x} p^X (1-p)^{n-x} & E(X)=np & {{n= \text{independent trials}} \\
{p= \text{probability of success}}} \\ \hline
{\text{Negative} \\ \text{Binomial}} & P(X=k)=\binom{k+r-1}{r-1} p^r (1-p)^k & \frac {r(1-p)}{p} & {{r= \text{No. of successes}} \\ {k= \text{No. of failures until the } r^{th} \text{ success.}} \\ {p= \text{probability of success}} }\\ \hline
\text{Geometric} & P(X=x)=(1-p)^{x-1} p & E(X)=\frac {1}{p} & \\ \hline
\text{Uniform} & P(X=x)=\frac {1}{n} & \frac {(n+1)}{2} & \\ \hline
\text{Hypergeometric} & P(X=x)=\frac {\binom{m}{x} \binom{N-m}{n-x}}{\binom{N}{n}} & E(X)=\frac {nm}{N} & { {n= \text{sample size}} \\ {N= \text{population size}} \\ {m= \text{successes}} \\ {N-m= \text{failures}} } \\ \hline
\text{MultinomiaL} & \frac {n!}{x_1 !,\dots,x_k !} p_1^{x_1}\dots p_k^{x_k} & E(X_i)=np_i & { {n= \text{independent trials}} \\
{p= \text{probability of success}}}
\end{array}} $$

Example:

Suppose that, on average, an insurance company receives five claims every half-year. Calculate the expected number of claims in 3 years.

Solution:

Let \(X\) be the number of claims received. It follows that \(X\) is a Poisson random variable with mean, \(\lambda\).

If, on average, the insurance company receives five claims every half-year, then in three years, we can expect, on average:

$$ \begin{align*} \frac {5}{0.5}\times 3 &=30 \\
\Rightarrow \lambda &=30 \end{align*} $$

Thus, the expected number of claims in 3 years is:

$$ E(X)=\lambda=30 $$

Example:

A meteorologist has observed that Town A is twice as likely to experience three tornadoes as it is to experience five tornadoes in a year. The number of tornadoes each year in Town A follows a Poisson distribution with mean λ. Determine the expected number of tornadoes in Town A each year.

Solution

We know that,

$$ P(X=x)=\frac {\lambda^x e^{-\lambda}}{xǃ} $$

From the given information,

$$ \begin{align*} P(X=3) &=2P(X=5) \\
\frac {\lambda^3 e^{-\lambda}}{3ǃ} &=2 \times \frac {\lambda^5 e^{-\lambda}}{5ǃ} \\
\lambda & =\sqrt{10}\approx 3.1623 \end{align*} $$

Thus, the expected number of tornadoes in Town A each year is approximately 3 tornadoes.

Expected Value of Continuous Random Variables

Now, if \(X\) is a continuous random variable with pdf \(f(x)\), then the expected value (or mean) of \(X\) is given by:

$$ E\left(X\right)=\int_{-\infty}^{\infty}{xf(x)dx} $$

Example: Expected Return of a Continuous Random Variable

A continuous random variable \(X\) has a probability density function (pdf) given by:

$$ f\left(x\right)= \left\{ \begin{matrix} \frac {x}{2}, & 0 \lt x \lt 2 \\ 0,& \text{otherwise} \end{matrix} \right. $$

1. Find \(E(X)\).
2. Find \(E(2X+10)\).

Solution

1. The expected value of a continuous random variable is given by:

   $$ \begin{align*}
   E\left(X\right) &= \int_{-\infty}^{\infty}{xf\left(x\right)dx} \\ & =\int_{0}^{2}{x.\frac{x}{2}dx=\left[\frac{x^3}{6}\right]_{x=0}^{x=2} }\\ & =\frac{8}{6}=\frac{4}{3}=1\frac{1}{3}
   \end{align*} $$

2. $$ \begin{align*} E\left(2X+10\right) & =\int_{0}^{2}{\left(2x+10\right)\times \frac{x}{2}dx} \\
   & =\int_{0}^{2}{\left(x^2+5x\right)=\left[\frac{x^3}{3}+\frac{5x^2}{2}\right]_{x=0}^{x=2}} \\ & =12.667 \end{align*} $$

The Mode

The mode of the random variable \(X\) is defined to be the value or values of the support of the random variable that maximizes the probability mass/density function.

Mode of Discrete Random Variables

Let \(X\) be a discrete random variable with probability mass function, \(p(x)\).

The mode of \({X}\) is the value \(x\), which is most likely to occur, with probability, \(p(x)\).

To find the mode for a discrete random variable, we list out all possible values of the random variable \(X\) and their corresponding probabilities \(p(x)\) and then identify the value \(x\) that has the highest probability \(p(x)\), which gives the mode of \(X\).

Example: Calculating the Mode of Discrete Random Variable

In an experiment of rolling two dice simultaneously, the following probabilities are obtained:

$$ \begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c}
X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline
p(x) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{2}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36}
\end{array} $$

What is the mode of the probability distribution of the sum of the two dice?

Solution

As shown in the table below, the most likely value is 7 with a probability of \(\frac{6}{36}\), so the mode is= 7.

A random variable can have more than one mode.

Example:

$$ \begin{array}{c|c}
X & P(X=x) \\ \hline
1 & 0.25 \\ \hline
3 & 0.2 \\ \hline
5 & 0.25 \\ \hline
8 & 0.3
\end{array} $$

In this case, we have two modes: 1 and 5.

The following are the common types of modes:

• Unimodal: A dataset with only one mode.
• Bimodal: A dataset with two modes.
• Multimodal: A dataset with more than two modes.

Mode of Continuous Random Variables

Let \(X\) be a continuous random variable with probability density function, \(f(x)\).

The mode of \(X\) is the value, \(x\), at which the probability density function, \(f(x)\), is at a maximum. i.e., the value of \(x\) is obtained by solving finding the first derivative of the function, \(f(x)\), equating it to 0, and solving for \(x\). This gives the mode of the continuous random variable, \(X\).

$$ f^\prime\left(x\right)=0 $$

It is a value that is most likely to lie within the same interval as the outcome.

$$ f^\prime\left(x\right)=0 $$

Remember: For the case of continuous random variables, the probability of a specific value occurring is zero, i.e., \(P\left(X=k\right)=0\) and the mode is a specific value.

Example: Mode of a Continuous Random Variable

Given the following probability density function of a continuous random variable, find the mode of the distribution.

$$ f\left(x\right)= \left\{ \begin{matrix} -x^2+2x-\frac{1}{6}, & 0\lt x \lt2 \\ 0, & \text{otherwise} \end{matrix} \right. $$

Solution

To find the mode, we have to find the maximum of f(x), by calculating the first derivative and setting that value equal to zero, as shown below:

$$ f^\prime(x) = -2x + 2 = 0 $$

Solving for \(x\), we have

$$ \begin{align*} -2x & =-2 \\ x&=1 \end{align*} $$

Therefore, the mode of the distribution is 1.

Percentiles and Median

The p^th percentile of a distribution is the value of \(x\) for which \(P(X\le x)\) is greater than or equal to \(\frac{p}{100}\) and \(P(X\geq x)\) is less than or equal to \(1-\frac{p}{100}\) which also means that \(P(X\geq x)\) is greater than or equal to \(\frac{p}{100}\).

Mathematically, this can be represented as,

$$ P\left(X\le x\right)=P\left(X\geq x\right)\geq\frac{p}{100} $$

Note that,

$$ P\left(X\geq x\right)=F(x) $$

The 25^th percentile of the continuous random variable \(X\) is defined to be the value of the random variable which is greater than 25% of the other values of the random variable.

Let us denote the 25^th percentile by \(p_{0.25}\) so that,

$$ \Pr{\left(X \lt p_{0.25}\right)}=0.25 $$

Recall that

$$ \int_{-\infty}^{p_{0.25}}{f_X(x)}dx=0.25 $$

Similarly, the 50^th percentile of the continuous random variable \(X\) is defined to be the value of the random variable which is greater than 50% of the other values of the random variable.

If we denote the 50^th percentile by \(p_{0.50}\) so that,

$$ \Pr{\left(X \lt p_{0.50}\right)}=0.50 $$

Recall that,

$$ \int_{-\infty}^{p_{0.50}}{f_X(x)}dx=0.50 $$

The 50^th percentile is referred to as the median.

Example: Calculating the Median of Discrete Random Distribution

Given the following probability density function of a discrete random variable, calculate the median of the distribution:

$$ p\left(x\right)= \left\{ \begin{matrix} 0.2, & x=1,4 \\ 0.3, & x=2,3, \end{matrix} \right. $$

Solution

The median of the distribution above is 2 because;

$$ P\left(X\le2\right)=P\left(X=1\right)+P\left(X=2\right)=0.2+0.3=0.5 $$

and,

$$ P\left(X\geq2\right)=P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)=0.3+0.3+0.2=0.8 $$

Example: Calculating the Median of a Continuous Distribution

Given the following probability density function of a continuous random variable, find the median of the distribution.

$$ f\left(x\right)= \left\{ \begin{matrix} -x^2+2x-\frac{1}{6},& 0 \lt x \lt 2 \\ 0, & \text{Otherwise} \end{matrix} \right. $$

Solution

We need,

$$ \begin{align*}
P\left(x\le m\right) & =0.5 \\
\Rightarrow\int_{0}^{m}\left(-x^2+2x-\frac{1}{6}\right)dx & =0.5 \\
=\left[-\frac{x^3}{3}+x^2-\frac{1}{6}x\right]_{x=0}^{x=m} &=0.5\\
=-\frac{m^3}{3}+m^2-\frac{1}{6}m & =0.5 \\
\Rightarrow m &=1 \end{align*} $$

Therefore, the median of the random variable is 1.

Example: Calculating the 75^th Percentile percentile from a Discrete Random Variable

Given the following probability density function of a discrete random variable, calculate the 75^th percentile of the distribution:

$$ f\left(x\right)=\left\{ \begin{matrix} 0.2, & x=1,4 \\ 0.3, & x= 3,4 \end{matrix} \right. $$

Solution

We need:

$$ P (X \le p_{0.75}) = .75 $$

So,

$$ \begin{align*}
P (X \le 3)& = P (X = 1) + P (X = 2) + P (X = 3) \\ & =0.2 + 0.3 + 0.2 = 0.7 \lt 0.75 \\
P (X \le 4)& = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) \\ & = 0.2 +0.3 + 0.2 + 0.3 = 1.0 \gt 0.75 \\
P (X \geq 4)& = P (X= 4)=0.2 \lt 0.25 = 1 – \left(\frac{75}{100}\right)
\end{align*} $$

So, the 75^th percentile of the distribution is 4.

In general, the p^th percentile of a continuous distribution can be defined as the value of \(c\) for which,

$$ \int_{-\infty}^{C}{f(x)dx}=\frac{p}{100} $$

Example: Calculating the 75^th Percentile from a Continuous Random Variable

Given the following probability density function of a continuous random variable, find the 25^th Percentile:

$$ f\left(x\right)=\left\{ \begin{matrix} -x^2+2x-\frac {1}{6}, & 0 \lt x \lt 2 \\ 0, & \text{otherwise} \end{matrix} \right. $$

Solution

We need:

$$ \begin{align*}
P\left(x\lt c\right) &=0.25 \\
\Rightarrow\int_{0}^{c}{\left(-x^2+2x-\frac{1}{6}\right)dx} & =\frac{25}{100}=0.25 \\
\left[-\frac{x^3}{3}+2x-\frac{1}{6}x\right]_{x=0}^{x=c}& =0.25 \\
-\frac{c^3}{3}+c^2-\frac{1}{6}c & =0.25 \\
c & =0.69\end{align*} $$

Higher Moments

The variance of \(X\) is sometimes referred to as the second moment of X about the mean. The third moment of X is referred to as the skewness, and the fourth moment is called kurtosis. In this reading, we will look at skewness and kurtosis, while variance will be covered in the next reading.

In general, the m^th moment of X can be calculated from the following formula:

$$ \it{m^{th} \text{ moment } \left(X\right)= \int_{-\infty}^{\infty}(x-\mu)^m f(x)dx } $$

This means that

$$ \text{Skew}\left(X\right)=\int_{-\infty}^{\infty}(x-\mu)^3 f(x)dx $$

and,

$$ \text{Kurtosis}\left(X\right)=\int_{-\infty}^{\infty}(x-\mu)^4 f(x)dx $$

Example: Calculating the Skewness

Given the following probability density function of a continuous random variable:

$$ f\left(x\right)= \left\{ \begin{matrix} \frac{x}{2}, & 0 \lt x \lt 2 \\ 0, & \text{otherwise} \end{matrix} \right. $$

Calculate the Skewness:

Solution

The skewness is calculated as the third moment of \(X\). More specifically,

$$ \it{\text{Skewness}\left(X\right)=\int_{-\infty}^{\infty}(x-\mu)^3 f(x)dx} $$

Now,

$$ \mu=\int_{-\infty}^{\infty}{xf(x)dx}=\int_{0}^{2}{x\times\frac{x}{2}\times d x=\left[\frac{x^3}{6}\right]_{x=0}^{x=2}=\frac{8}{6}=\frac{4}{3}} $$

Therefore,

$$ \begin{align*}
\it{Skew\left(X\right)} &=\it{\int_{-\infty}^{\infty}(x-\mu)^3 f(x)dx } \\
& =\int_{0}^{2}{\left(x-\frac{4}{3}\right)^3\times\frac{x}{2}\times d x} \\
& =\int_{0}^{2}\left(\frac{x^4}{2}-2x^3+\frac{8x^2}{3}-\frac{32x}{27}\right)dx \\
& =\left[\frac{x^5}{10}-\frac{x^4}{2}+\frac{8x^3}{9}-\frac{16x^2}{27}\right]_{x=0}^{x=2} \\ & =-\frac{8}{135}
\end{align*} $$

Question

A health insurance policy pays an individual $150 per day for up to 2 days of hospitalization and $50 per day for each day of hospitalization thereafter. The number of days of hospitalization, Y, is a discrete random variable with the following probability function:

$$ p\left(Y=j\right)=\left\{ \begin{matrix} \frac{1}{10}(5-j) & \text{for }j=1,2,3,4 \\ 0 & \text{elsewhere} \end{matrix} \right. $$

Calculate the expected payment for hospitalization under this policy.

1. $150
2. $260
3. $130
4. $100
5. $220

Solution

The correct answer is B.

To calculate the expected payment, we need to find \(E(Y)\).We can use the following table:

$$ \begin{array}{c|c|c|c|c}
\text{Days (j)} & 1 & 2 & 3 & 4 \\ \hline
\text{Payment} & \$150 & \$300 & \$350 & \$400 \\ \hline
P(Y=j) & =\frac{1}{10}\times(5-1)=\frac{4}{10} & \frac{3}{10} & \frac{2}{10} & \frac{1}{10}
\end{array} $$

$$ E\left(Y\right)=150\left(\frac{4}{10}\right)+300\left(\frac{3}{10}\right)+350\left(\frac{2}{10}\right)+400\left(\frac{1}{10}\right)=\$260 $$

Alternatively, we can compute the mean as follows:

$$ \begin{align*}
E\left(Y\right) &=\sum_{j=1}^{2}{150j\left(\frac{1}{10}\right)\left(5-j\right) }\\ &+\sum_{j=3}^{4}\left(200+50j\right)\left(\frac{1}{10}\right)\left(5-j\right) \\
& =150\times1\times\left(\frac{1}{10}\right)\times4+150\times2\times\left(\frac{1}{10}\right)\times3 \\ & +\left(200+50\left(3\right)\right)\left(\frac{1}{10}\right)\times2+ \left(200+50\left(4\right)\right)\left(\frac{1}{10}\right)\times1 \\
& =60+90+70+40 = \$260 \end{align*} $$

Therefore, the expected benefit payment from the insurance policy is approx. $260.

Learning Outcome

Topic 2.c: Univariate Random Variables – Explain and calculate expected value, mode, median, Percentile, and higher moments.