Math Problem Statement

If X is uniform over (0,1), calculate E(Xn) and Var(Xn)

Solution

Given that XX is uniformly distributed over (0,1)(0, 1), and we want to calculate E(Xn)E(X^n) and Var(Xn)\text{Var}(X^n), we can proceed as follows:

1. Expectation E(Xn)E(X^n)

For a random variable XX uniformly distributed over (0,1)(0, 1), the probability density function (PDF) is:

fX(x)=1for0x1.f_X(x) = 1 \quad \text{for} \quad 0 \leq x \leq 1.

To calculate E(Xn)E(X^n), we use the definition of expectation:

E(Xn)=01xnfX(x)dx=01xndx.E(X^n) = \int_0^1 x^n f_X(x) \, dx = \int_0^1 x^n \, dx.

This is a simple integral:

E(Xn)=[xn+1n+1]01=1n+1.E(X^n) = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 = \frac{1}{n+1}.

Thus, the expectation is:

E(Xn)=1n+1.E(X^n) = \frac{1}{n+1}.

2. Variance Var(Xn)\text{Var}(X^n)

Variance is defined as:

Var(Xn)=E((Xn)2)(E(Xn))2.\text{Var}(X^n) = E((X^n)^2) - (E(X^n))^2.

a. First, calculate E((Xn)2)E((X^n)^2):

E((Xn)2)=E(X2n)=01x2ndx.E((X^n)^2) = E(X^{2n}) = \int_0^1 x^{2n} \, dx.

This integral evaluates similarly to the expectation of XnX^n:

E(X2n)=12n+1.E(X^{2n}) = \frac{1}{2n+1}.

b. Now, calculate the variance:

Var(Xn)=12n+1(1n+1)2.\text{Var}(X^n) = \frac{1}{2n+1} - \left( \frac{1}{n+1} \right)^2.

Simplifying the expression:

Var(Xn)=12n+11(n+1)2.\text{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}.

Final Results:

  • E(Xn)=1n+1E(X^n) = \frac{1}{n+1}
  • Var(Xn)=12n+11(n+1)2\text{Var}(X^n) = \frac{1}{2n+1} - \frac{1}{(n+1)^2}.

Would you like a more detailed explanation or have any questions?

Here are five related questions:

  1. How does the uniform distribution affect the expectation calculation?
  2. What changes if XX is uniformly distributed over (a,b)(a, b) instead of (0,1)(0, 1)?
  3. How would the expectation change if XX followed a different distribution, like exponential?
  4. What is the significance of calculating Var(Xn)\text{Var}(X^n)?
  5. How does the integral method change for non-uniform distributions?

Tip: For uniform distributions, integrals involving powers of XX often follow simple patterns, making calculations straightforward.

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Math Problem Analysis

Mathematical Concepts

Probability
Uniform Distribution
Expectation
Variance

Formulas

E(X^n) = ∫_0^1 x^n dx = 1 / (n + 1)
Var(X^n) = E(X^2n) - (E(X^n))^2 = (1 / (2n + 1)) - (1 / (n + 1))^2

Theorems

Definition of Expectation
Definition of Variance

Suitable Grade Level

Undergraduate