Mathematical expectation: simple random variables (Page 2/4)

Applied probability Page 2 / 4

E [X] = \sum_{i = 1}^{n} t_{i} P (X = t_{i}) = \sum_{i = 1}^{n} t_{i} p_{i}

Note that the expectation is determined by the distribution. Two quite different random variables may have the same distribution, hence the same expectation.Traditionally, this average has been called the mean , or the mean value , of the random variable X .

Some special cases

Since $X = a I_{E} = 0 I_{E^{c}} + a I_{E}$ , we have $E [a I_{E}] = a P (E)$ .
For X a constant c , $X = c I_{Ω}$ , so that $E [c] = c P (Ω) = c$ .
If $X = \sum_{i = 1}^{n} t_{i} I_{A_{i}}$ then $a X = \sum_{i = 1}^{n} a t_{i} I_{A_{i}}$ , so that
$E [a X] = \sum_{i = 1}^{n} a t_{i} P (A_{i}) = a \sum_{i = 1}^{n} t_{i} P (A_{i}) = a E [X]$

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Figure 1 is a drawing of the moment of a probability distribution about the origin. The expected value of X, E[X], is equal to the sum of the moments, which is equal to the center of mass. The drawing shows one major horizontal line split in half by one major vertical line. As a title, the top of the drawing reads Negative Moments to the left of the vertical line, and Positive Moments to the right, which are meant to distinguish the arrows and labels in the drawing. On the horizontal line are five black dots, two to the left of the vertical line and three to the right. Below the corresponding dots are the corresponding labels: t_1, t_2, t_3, t_4, and t_5. Above the black dots are the following labels: p_1, p_2, p_3, p_4, and p_5. Above the horizontal line is another smaller horizontal line with arrows pointing in both directions. The label for the arrow pointing to the left is t_2 p_2, and the label for the arrow on the left is t_3 p_3. A longer horizontal line sits further up on the drawing, which also has arrows pointing in both directions. and intersects the same vertical line. The arrows are approximately twice as long as the two arrows below. The label for the arrow pointing to the left is t_1 p_1, and the label for the arrow to the left is t_4 p_4. finally, there is one horizontal line extending only to the right of the vertical line, with an arrow pointing to the right. This line is longer in this direction than any of the arrows that sit below it pointing to the right. The arrow is labeled t_5 p_5. — Moment of a probability distribution about the origin.

Mechanical interpretation

In order to aid in visualizing an essentially abstract system, we have employed the notion of probability as mass. The distribution induced bya real random variable on the line is visualized as a unit of probability mass actually distributed along the line. We utilize the mass distribution to give an important andhelpful mechanical interpretation of the expectation or mean value. In Example 6 in "Mathematical Expectation: General Random Variables", we give an alternate interpretation in terms of mean-square estimation.

Suppose the random variable X has values ${t_{i} : 1 \leq i \leq n}$ , with $P (X = t_{i}) = p_{i}$ . This produces a probability mass distribution, as shown in Figure 1, with point mass concentration in the amount of p _i at the point t _i . The expectation is

\sum_{i} t_{i} p_{i}

Now $| t_{i} |$ is the distance of point mass p _i from the origin, with p _i to the left of the origin iff t _i is negative. Mechanically, the sum of the products $t_{i} p_{i}$ is the moment of the probability mass distribution about the origin on the real line. From physical theory, thismoment is known to be the same as the product of the total mass times the number which locates the center of mass. Since the total mass is one, the mean value is the location of the center of mass . If the real line is viewed as a stiff, weightless rod with point mass p _i attached at each value t _i of X , then the mean value μ _X is the point of balance. Often there are symmetries in the distribution which make it possible todetermine the expectation without detailed calculation.

The number of spots on a die

Let X be the number of spots which turn up on a throw of a simple six-sided die. We suppose each number is equally likely. Thus the values are the integers one through six,and each probability is 1/6. By definition

E [X] = \frac{1}{6} \cdot 1 + \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot 3 + \frac{1}{6} \cdot 4 + \frac{1}{6} \cdot 5 + \frac{1}{6} \cdot 6 = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) = \frac{7}{2}

Although the calculation is very simple in this case, it is really not necessary. The probability distribution places equal mass at each of the integer values one throughsix. The center of mass is at the midpoint.

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A simple choice

A child is told she may have one of four toys. The prices are $2.50. $3.00, $2.00, and $3.50, respectively. She choses one, with respective probabilities 0.2, 0.3, 0.2, and 0.3 of choosing the first,second, third or fourth. What is the expected cost of her selection?

E [X] = 2.00 \cdot 0.2 + 2.50 \cdot 0.2 + 3.00 \cdot 0.3 + 3.50 \cdot 0.3 = 2.85

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For a simple random variable, the mathematical expectation is determined as the dot product of the value matrix with the probability matrix. This is easilycalculated using MATLAB.

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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