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Chapter 4
Expected Value
Definition
Let X be a random variable with probability distribution f(x). The mean, or expected value, of X is:
For a discrete distribution
E[X] = \sum\limits_x xf(x)
For a continuous distribution:
E[X] = \int\limits_{-\infty}^{\infty} xf(x)dx
Given \{1, 2, 3, 3, 5\}, the mean is:
{1+2+3+3+5 \over 5} = 2.8
$$f(x) = \begin{cases} {1\over5} & x=1 \ {1\over5} & x=2 \ {2\over5} & x=3 \ {1\over5} & x=5 \ \end{cases}$$
\sum\limits_x xf(x) = {1\over5}(1) + {1\over5}(2) + {1\over5}(3) + {1\over5}(5) = 2.8
Example
The probability distribution of a discrete random variable X is:
f(x) = {3 \choose x}\left({1 \over 4}\right)^x\left({3\over4}\right)^{3-x}, x \in \{0, 1, 2, 3\}
Find E[X]:
$$f(x) =
\begin{cases}
0 & x=0 \
0.422 & x=1 \
0.14 & x=2 \
{1\over64} & x=3
\end{cases}$$
E[X] = \sum\limits_x x {3 \choose x}\left({1\over4}\right)^x \left({3\over4}\right)^{3-x}
E[X] = 0(0)+ 0.422(1) + 0.14(2) + {1\over64}(3) = 0.75
Example
Let X be the random variable that denotes the life in hours of a certain electronic device. The PDF is:
$$f(x) =
\begin{cases}
{20000\over x^3} & x > 100 \
0 & elsewhere
\end{cases}$$
Find the expected life of this type of device:
E[X] = \int\limits_{-\infty}^{\infty} xf(x)dx = \int\limits_{100}^{\infty}x{20000 \over x^3}dx = 200 \text{[hrs]}
Note:
E[x^2] = \int\limits_{\infty}^{\infty}x^2f(x)dx
Properties of Expectations
E(b) = b
Where b is a constant
E(aX) = aE[X]
Where a is a constant
E(aX + b) aE[X] + b
E[X + Y] = E[X] + E[Y]
Where X and Y are random variables
Example
Given:
$$f(x) = \begin{cases}
{x^2\over3} & -1 < x < 2 \
0 & \text{elsewhere}
\end{cases}$$
Find the expected value of Y = 4X + 3:
E[Y] = E[4X + 3] = 4E[X] + 3
E[X] = \int\limits_{-1}^{3} {X^3 \over 3}dx = {1\over12}X^4 \Big|_{-1}^{3}={5\over4}
Variance of a Random Variable
The expected value/mean is of special importance because it describes where the probability distribution is centered. However, we also need to characterize the variance of the distribution.
Definition
Let X be a random variable with probability distribution, f(x), and mean, \mu. The variance of X is given by:
\text{Var}[X] = E[(X-\mu)^2]
Which is the average squared distance away from the mean. This simplifies to:
\text{Var}[X] = E[X^2] - E[X]^2
Note: Generally,
E[X^2] \ne E[X]^2
The standard deviation, \sigma, is given by:
\sigma = \sqrt{\text{Var}[X]}
Note: The variance is a measure of uncertainty (spread) in the data.
Example
The weekly demand for a drinking water product in thousands of liters from a local chain of efficiency stores is a continuous random variable, X, having the probability density:
$$F(x) = \begin{cases}
2(x-1) & 1 < x < 2 \
0 & \text{elsewhere}
\end{cases}$$
Find the expected value:
E[X] = \int\limits_1^2 x (2(x-1)) dx = 2\int\limits_1^2 (x^2 - x)dx
E[X] = 2\left[{1\over3}x^3 - {1\over2}x^2 \Big|_1^2 \right] = {5\over3}
Find the variance:
\text{Var}[X] = E[X^2] - E[X]^2
E[X^2] = \int\limits_1^2 2x^2(x-1)dx = 2\int\limits_1^2 (x^3 - x^2)dx
E[X^2] = {17\over6}
\text{Var}[X] = {17\over6} - \left({5\over3}\right)^2 = {1\over18}
Find the standard deviation:
\sigma = \sqrt{\text{Var}[X]} = {1\over3\sqrt{2}} = {\sqrt{2}\over6}
Example
The mean and variance are useful when comparing two or more distributions.
| Plan 1 | Plan 2 | |
|---|---|---|
| Avg Score Improvement | +17 |
+15 |
| Standard deviation | \pm8 |
\pm2 |
Theorem
If X has variance, \text{Var}[X], then \text{Var}[aX + b] = a^2\text{Var}[X].
Example
The length of time, in minutes, for an airplane to obtain clearance at a certain airport is a random variable, Y = 3X - 2, where X has the density:
$$F(x) = \begin{cases}
{1\over4} e^{x/4} & x > 0 \
0 & \text{elsewhere}
\end{cases}$$
E[X] = 4
\text{Var}[X] = 16
Find E[Y]:
E[Y] = E[3X-2] = 3E[X] - 2 = 10
\text{Var}[Y] = 3^2\text{Var}[X] = 144
\sigma = \sqrt{\text{Var}[Y]} = 12
The Exponential Distribution
The continuous random variable, X, has an exponential distribution with parameter \beta if its density function is given by:
$$F(x) = \begin{cases}
{1\over\beta}e^{-x/\beta} & x > 0 \
0 & \text{elsewhere}
\end{cases}$$
Where \beta > 0.
E[X] = \beta
E[X] = \int\limits_0^{\infty} x{1\over\beta}e^{-x/\beta} dx
Aside:
\Gamma(Z) = \int\limits_0^\infty x^{Z - 1}e^{-x}dx
Where \Gamma(Z) = (Z - 1)!
E[X] = \beta \int\limits_0^\infty \left({x\over\beta}\right)^{(2-1)} e^{-x/\beta} \left({dx\over\beta}\right) = \beta\Gamma(2)
E[X] = \beta(2-1)! = \beta
\text{Var}[X] = E[X^2] - E[X]^2
E[X^2] = \int\limits_0^\infty x^2{1\over\beta}e^{-x/\beta}dx = \beta^2 \int\limits_0^\infty \left({x\over\beta}\right)^{(2-1)} e^{-x/\beta} \left({dx\over\beta}\right)
E[X^2] = \beta^2\Gamma(3) = 2\beta^2
\text{Var}[X] = 2\beta^2 - \beta^2 = \beta^2
Application
Reliability analysis: the time to failure of a certain electronic component can be modeled by an exponential distribution.
Example
Let T be the random variable which measures the time to failure of a certain electronic component. Suppose T has an exponential distribution with \beta = 5.
$$F(x) = \begin{cases} {1\over5}e^{-x/5} & x > 0 \ 0 & \text{elsewhere} \end{cases}$$
If 6 of these components are in use, what is the probability that exactly 3 components are still functioning at the end of 8 years?
What is the probability that an individual component is still functioning after 8 years?
P(T > 8) = \int\limits_8^\infty {1\over5}e^{-x/5}dx \approx 0.2
{6 \choose 3}(0.2)^3(0.8)^3 = 0.08192
>>> from math import comb
>>> comb(6,3) * 0.2**3 * 0.8**3
0.08192000000000003
The Normal Distribution
The most important continuous probability distribution in the field of statistics is the normal distribution. It is characterized by 2 parameters, the mean, \mu, and the variance, \sigma^2.
\text{mean} = \text{median} = \text{mode}
F(x|\mu,\sigma^2) = {1 \over \sqrt{2\pi} \sigma^2} e^{\left({1 \over 2\sigma^2}(x-\mu)^2\right)}
E[X] = \mu
\text{Var}[X] = \sigma^2
For a normal curve:
P(x_1 < x < x_2) = \int\limits_{x_1}^{x_2} F(x)dx
Definition
The distribution of a normal variable with mean 0 and variance 1 is called a standard normal distribution.
The transformation of any random variable, X into a standard normal variable, Z:
Z = {X - \mu \over \sigma}
Example
Given a normal distribution with mean \mu = 30 and standard deviation, \sigma = 6, find the normal curve area to the right of x = 17.
Transform to standard normal.
Z = {17 - 30 \over 6} = -2.16
That is, x = 17 on a normal distribution with \mu = 30 and \sigma = 6 is equivalent to Z=-2.16 on a normal distribution with \mu = 0 and \sigma = 1.
P(X > 17) = P(Z > -2.16)
P(Z > -2.16) = 1 -P(Z \le -2.16) = 0.9846
>>> from scipy.stats import norm
>>> norm.cdf(-2.16)
0.015386334783925445
Example
The finished inside diameter of a piston ring is normally distributed with mean, $\mu = 10$[cm], and standard deviation, $\sigma = 0.03$[cm].
What is the probability that a piston ring will have inside diameter between 9.97[cm] and 10.03[cm]?
Z_1 = {9.97 - 10 \over 0.03} = -1
Z_2 = {10.03 - 10 \over 3} = 1
P(9.97 < x < 10.03) = 0.68
>>> from scipy.stats import norm
>>> norm.cdf(1) - norm.cdf(-1)
0.6826894921370859