Question

Let X1, Xn be i.i.d. random variables with probability density function,

fx (x) = { 1/0 0 0 < x < 0 otherwise.

(a) Let X₁, X denote a bootstrap sample and let X = n

Find: E(XX1, , Xn), V(XX1, , Xn), E(X), V(X₂).

Hint: Law of total expectation: E(X)= E(E(XY)).

Law of total variance: V(X) = E(V(XY))+ (E(X|Y)).

Sample variance, i.e. s² = (X; - X)2 is an unbiased estimator of population variance.

Answer 1

The expected value of the assumed probability density function is 3/4 and the variance is 3/80

How to solve for E(x) and V(x)?

The probability density function is not properly stated.

So, I will provide a general explanation on how to determine E(x) and V(x) of a probability density function

Assume that the probability density function is given as:

$f(x)= \left[\begin{array}{ccc}3x^2 &0 \le x \le 1\\0&Otherwise\end{array}\right$

The expression E(x) is calculated using:

$E(x) = \int\limits^a_b {x \cdot f(x)} \, dx$

So, we have:

$E(x) = \int\limits^1_0 {x \cdot 3x^2 \ d(x)$

Evaluate the product

$E(x) = \int\limits^1_0 { 3x^3 \ d(x)$

Integrate

$E(x) = \frac34x^4 |\limits^1_0$

Expand

$E(x) = \frac34(1-0)^4$

Evaluate

$E(x) = \frac34$

The variance is then calculated using:

$Var(x) = E(x^2) - (E(x))^2$

Calculate $E(x^2)$ using:

$E(x^2) = \int\limits^a_b {x^2 \cdot f(x)} \, dx$

This gives

$E(x^2) = \int\limits^1_0 {x^2 \cdot 3x^2} \, dx$

Expand

$E(x^2) = \int\limits^1_0 {3x^4} \, dx$

Integrate

$E(x^2) = \frac{3}{5}x^5|\limits^1_0$

Expand

$E(x^2) = \frac{3}{5}(1 - 0)^5$

Evaluate

$E(x^2) = \frac{3}{5}$

Substitute $E(x^2) = \frac{3}{5}$ and $E(x) = \frac34$ in $Var(x) = E(x^2) - (E(x))^2$

$Var(x) = \frac 35 - (\frac{3}{4})^2$

Evaluate the exponent

$Var(x) = \frac 35 - \frac{9}{16}$

Evaluate the difference

$Var(x) = \frac{3}{80}$

Hence, the variance of the probability density function is 3/80

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https://brainly.com/question/15858152

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