Expected value (or mean value) of a random variable is a key concept in probability and statistics that represents the average or long-term value of the outcomes of a random process. It's the weighted average of all possible values a random variable can take, with each value weighted by its probability of occurrence.
For a discrete random variable X, the expected value E(X) is calculated as:
E(X) = \sum_{i=1}^{n} x_i P(x_i)
Where:
- xi represents each possible value that the random variable X can take.
- P(xi) is the probability that X takes the value xi.
For a continuous random variable, the expected value is calculated using an integral:
E(X) = \int_{-\infty}^{\infty} x f(x) \, dx
where f(x) is the probability density function (PDF) of the random variable.
Table of Content
Properties of Expected Value
Expected value (or mean) has several important properties that make it useful for probability theory and statistics. Here are the key properties of expected value:
Linearity of Expectation
The expected value of the sum (or difference) of two random variables is equal to the sum (or difference) of their expected values. This property holds regardless of whether the variables are independent or not.
- E(X + Y) = E(X) + E(Y)
- E(X - Y) = E(X) - E(Y)
For any constant ccc and random variable X:
E(cX) = cE(X)
Example: If E(X) = 5 and E(Y) = 3, then E(X + Y) = 5 + 3 = 8.
Expectation of a Constant
The expected value of a constant ccc is just c. This makes sense because if a random variable always takes the same value, the "average" or "mean" of that constant value is simply the value itself.
E(c) = c
Example: If X = 4 with certainty, then E(X) = 4.
Multiplying by a Constant
The expected value of a constant multiplied by a random variable is equal to the constant multiplied by the expected value of the random variable.
E(cX) = cE(X)
This shows how scaling affects the mean of a random variable.
Example: If E(X) = 7 and c = 2, then E(2X) = 2 × 7 = 14.
Expected Value of the Product (Independence)
If X and Y are independent random variables, the expected value of their product is the product of their expected values:
E(XY) = E(X) ⋅ E(Y)
This property applies only if X and Y are independent.
Example: If E(X) = 5, E(Y) = 3, and X and Y are independent, then E(XY) = 5 × 3 = 15.
Non-Negativity of Expectation
If a random variable X is always non-negative (i.e., X ≥ 0), then its expected value is also non-negative.
X ≥ 0 ⟹ E(X) ≥ 0
Example: For a random variable that represents a non-negative quantity, such as the number of customers arriving at a store, E(X) ≥ 0.
Additivity over Partitions (Law of Total Expectation)
If you partition the sample space into disjoint events A1, A2, . . . , An, the expected value can be decomposed as a weighted sum, where each weight is the probability of the corresponding event:
E(X) = P(A1)E(X∣A1) + P(A2)E(X∣A2) + . . . + P(An)E(X∣An)
This is often used in scenarios where the probability distribution of X depends on some condition or event.
Monotonicity of Expectation
If X ≤ Y for all possible values, then the expected value of X is less than or equal to the expected value of Y:
X ≤ Y ⟹ E(X) ≤ E(Y)
This property shows that if one random variable is always smaller than another, its expected value will also be smaller.
Expected Value and Variance
The expected value is related to the variance of a random variable. Specifically, the variance is a measure of how much the values of a random variable deviate from its expected value. The formula for variance is:
\text{Var}(X) = E(X^2) - [E(X)]^2
Solved Problems
Problem 1: Given E(X) = 3 and E(Y) = 5, find E(4X + 2Y).
Solution:
Using the linearity of expectation:
E(4X + 2Y) = 4 E(X) + 2 E(Y) = 4 × 3 + 2 × 5 = 12 + 10 = 22The answer is 22.
Problem 2: If X = 5, find E(7 + X).
Solution:
Since the expected value of a constant is the constant itself:
E(7 + X) = E(7) + E(X) = 7 + 5 = 12The answer is 12.
Problem 3: Let E(X) = 4 and E(Y) = 6, and assume X and Y are independent. Calculate E(XY).
Solution:
Using the property of the product of independent variables:
E(XY) = E(X)⋅ E(Y) = 4 × 6 = 24
The answer is 24.
Problem 4: If f(X) = 2X + 3, and E(X) = 4, find E(f(X)).
Solution:
E(f(X)) = E(2X + 3) = 2E(X) + E(3) = 2 × 4 + 3 = 8 + 3 = 11
The answer is 11.
Problem 5: Given X≤Y, and E(X) = 4 and E(Y) = 7, verify if the monotonicity property holds.
Solution:
Since X ≤ Y, by the monotonicity property: E(X) ≤ E(Y)
Thus, 4≤7, which is true.
Problem 6: If E(X) = 6, find E(3X).
Solution:
Using the scaling property of expectation:
E(3X) = 3 × E(X) = 3 × 6 = 18
The answer is 18.
Problem 7: If E(X) = 7 and E(Y) = 2, find E(2X−3Y).
Solution:
Using linearity of expectation:
E(2X−3Y) = 2E(X) − 3E(Y) = 2 × 7 −3 × 2 = 14−6 = 8The answer is 8.
Problem 8: Find E(5X + 2Y) when E(X) = 3 and E(Y) = 4.
Solution:
(5X + 2Y) = 5E(X) + 2E(Y) = 5 × 3 + 2 × 4 = 15 + 8 = 23
The answer is 23.
Practice Questions on Properties of Expected Value
You can download free worksheet on Properties of Expected Value for practicing various different questions with their answers from below:
Conclusion
In conclusion, the properties of expected value are essential for understanding how averages work in probability and statistics. These properties, like the ability to add or scale expected values easily, help us simplify calculations involving random events.
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