Expected Value Properties
Let X be a discrete random variable following a Uniform distribution from 1 to N, i.e., X ~ Uniform(1, N). This means X can take any integer value from 1 to N, each with probability 1/N.
Find:
- E[X²] (The expected value of X squared)
- E[XY], where Y is another random variable, independent of X, and also distributed as Y ~ Uniform(1, N).
Related Concepts
Hint
- For E[X²] (Average of X-squared):
- Remember how to calculate the average (expected value) of something: you take each possible value of that "something", multiply it by its probability, and add them all up. Here, "something" is X².
- So, E[X²] = (1² × P(X=1)) + (2² × P(X=2)) + ... + (N² × P(X=N)).
- For a Uniform(1, N) distribution, X can be any whole number from 1 to N, and each number has the same chance of occurring. What is this chance (probability) for each number? (Answer: 1/N).
- You will need to sum up the squares: 1² + 2² + ... + N². There's a handy formula for this sum: Σk=1N k² = N(N+1)(2N+1)/6.
- For E[XY] (Average of the product of X and Y):
- The problem states that X and Y are independent. This is a very important clue! When two random variables are independent, there's a special rule for the average of their product: E[XY] = E[X] × E[Y].
- So, you first need to find the average of X (E[X]). What is E[X] for a variable that is uniformly distributed from 1 to N? (Think about the average of numbers from 1 to N. Or use the formula: E[X] = (1+N)/2).
- Since Y has the exact same distribution as X (Uniform(1, N)), its average (E[Y]) will be the same as E[X].
- Once you have E[X] and E[Y], just multiply them.
Explanation: Expected Value Properties
Imagine we have a special spinner that can land on any whole number from 1 up to N. Each number has an equal chance of coming up. This is what "X ~ Uniform(1, N)" means. Let's use a familiar example: a fair 6-sided die. Here, N=6. The numbers are 1, 2, 3, 4, 5, 6, and each has a 1/6 chance.
- E[X²] (The average of the squared values): For our die (N=6):
- Possible outcomes for X: 1, 2, 3, 4, 5, 6.
- Squared outcomes (X²): 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36.
- Each squared outcome also has a 1/6 chance.
- To find the average of these squared outcomes:
E[X²] = (1×1/6) + (4×1/6) + (9×1/6) + (16×1/6) + (25×1/6) + (36×1/6)
= (1+4+9+16+25+36)/6 = 91/6 ≈ 15.17.
- E[XY] (The average of products of two independent rolls/spins): Imagine we have two such dice (or spinners), X and Y. They are "independent," meaning the result of one doesn't affect the other. We roll both, get a number from X and a number from Y, and multiply them. We want the average of this product if we did it many times.
- For our dice (N=6): First, what's the average roll for one die, E[X]? It's (1+2+3+4+5+6)/6 = 21/6 = 3.5.
- Since Y is an identical die, E[Y] is also 3.5.
- Because they are independent, a cool math trick says E[XY] = E[X] × E[Y].
- So, E[XY] = 3.5 × 3.5 = 12.25.
This question asks us to find these general formulas.
Given X ~ Uniform(1, N). This means X can take any integer value k from 1 to N, and the probability of X taking any specific value k, P(X=k), is 1/N.
1. Calculating E[X²]
The expected value (or average) of X² is found by summing up each possible value of X² multiplied by its probability. The possible values for X are k = 1, 2, ..., N. So the possible values for X² are k² = 1², 2², ..., N². The probability P(X=k) is 1/N for each k.
By definition: E[X²] = Σk=1N k² * P(X=k)
Since P(X=k) = 1/N for all k from 1 to N:
E[X²] = Σk=1N k² * (1/N)
We can take the constant (1/N) out of the sum:
E[X²] = (1/N) * Σk=1N k²
Now we need the formula for the sum of the first N squares: 1² + 2² + ... + N². This sum is given by the formula: Σk=1N k² = N(N+1)(2N+1)/6
Substitute this sum back into our equation for E[X²]:
E[X²] = (1/N) * [N(N+1)(2N+1)/6]
The 'N' in the (1/N) cancels out with the 'N' in the numerator of the sum formula:
E[X²] = (N+1)(2N+1)/6
This is our first answer.
2. Calculating E[XY]
We are given that X and Y are independent random variables. Both X and Y follow the same Uniform(1, N) distribution.
A very important property for independent random variables is that the expected value of their product is the product of their expected values: E[XY] = E[X] × E[Y] (This is true only because X and Y are independent).
First, let's find E[X]. For a discrete uniform distribution from 1 to N: E[X] = Σk=1N k * P(X=k) E[X] = Σk=1N k * (1/N) E[X] = (1/N) * Σk=1N k
We use the formula for the sum of the first N integers: 1 + 2 + ... + N. This sum is given by: Σk=1N k = N(N+1)/2
Substitute this sum back into the equation for E[X]: E[X] = (1/N) * [N(N+1)/2]
The 'N' cancels out:
E[X] = (N+1)/2
Since Y has the same distribution as X (Uniform(1, N)), its expected value E[Y] will be the same: E[Y] = (N+1)/2
Now, we can calculate E[XY] using the independence property:
E[XY] = E[X] × E[Y] E[XY] = [(N+1)/2] × [(N+1)/2]
E[XY] = (N+1)² / 4
This is our second answer.
Let's check with our N=6 (die roll) example:
- E[X²] for N=6:
Using the formula:E[X²] = (N+1)(2N+1)/6 = (6+1)(2*6+1)/6 = (7)(12+1)/6 = (7)(13)/6 = 91/6.
This matches our earlier manual calculation! - E[X] for N=6:
Using the formula:E[X] = (N+1)/2 = (6+1)/2 = 7/2 = 3.5.
This also matches. - E[XY] for N=6:
Using the formula:E[XY] = (N+1)² / 4 = (6+1)² / 4 = 7² / 4 = 49/4 = 12.25.
This also matches our earlier calculation (3.5 × 3.5 = 12.25).
Key Formulas and Concepts Used:
- Definition of Expected Value for g(X): For a discrete random variable X taking values xi with probability P(xi), E[g(X)] = Σ g(xi)P(xi).
- Discrete Uniform Distribution (1 to N): P(X=k) = 1/N for k = 1, 2, ..., N.
- Sum of first N integers: Σk=1N k = N(N+1)/2
- Sum of first N squares: Σk=1N k² = N(N+1)(2N+1)/6
- Property of Expectation for independent variables: If X and Y are independent, then E[XY] = E[X]E[Y].
These are fundamental results in probability and statistics. The term E[X²] is particularly important as it is used in calculating the variance of a random variable, which measures its spread or dispersion: Var(X) = E[X²] - (E[X])².
Challenge: Using the results for E[X] = (N+1)/2 and E[X²] = (N+1)(2N+1)/6, can you find the formula for the Variance, Var(X), for a discrete uniform distribution X ~ Uniform(1, N)? Recall the formula for variance: Var(X) = E[X²] - (E[X])².