Question

Let X, and Y be two independent random variables such that E(X)=2, Var(X)=4, E(Y)=7, and Var(Y)=1. Find: 1). E(3X+7) and Var(3X+7) 2). E(5X+2Y-2) and Var(5X+2Y-2).
(b) If two random variables have the joint density
Find the probability that

Answer 1

Answer on Question #45893 – Math – Statistics and Probability

Problem.

Let $X$, and $Y$ be two independent random variables such that $E(X) = 2$, $\operatorname{Var}(X) = 4$, $E(Y) = 7$, and $\operatorname{Var}(Y) = 1$.

Find:

1). $E(3X + 7)$ and $\operatorname{Var}(3X + 7)$

2). $E(5X + 2Y - 2)$ and $\operatorname{Var}(5X + 2Y - 2)$.

(b) If two random variables have the joint density

Find the probability that

Remark:

The part of b question is missed, so this is the answer to a question.

Solution:

1) The expectation operator is linear, so

$$E(3X + 7) = E(3X) + E(7) = 3E(X) + E(7) = 6 + 7 = 13.$$ $$\operatorname{Var}(X) = E(X^2) - \left(E(X)\right)^2 = 4, \text{ so } E(X^2) = \operatorname{Var}(X) + \left(E(X)\right)^2 = 8.$$ $$\begin{array}{l} \operatorname{Var}(3X + 7) = E((3X + 7)^2) - \left(E(3X + 7)\right)^2 = E((3X + 7)^2) - 169 = \\ = E(9X^2 + 42X + 49) - -169 = E(9X^2) + E(42X) + E(49) - 169 = \\ = 9E(X^2) + 42E(X) + E(49) - 169 = \\ \end{array}$$

$= 9 \cdot 8 + 42 \cdot 2 + 49 - 169 = 36$, by the definition of the variance.

Other method uses properties of variances:

$$\operatorname{Var}(3X + 7) = \operatorname{Var}(3X) = 3^2\operatorname{Var}(X) = 9\operatorname{Var}(X) = 9 \cdot 4 = 36$$

2) The expectation operator is linear, so

$$E(5X + 2Y - 2) = E(5X) + E(2Y) + E(-2) = 5E(X) + 2E(Y) - 2 = 10 + 14 - 2 = 22.$$ $$\operatorname{Var}(X) = E(X^2) - \left(E(X)\right)^2 = 4 \text{ and } \operatorname{Var}(Y) = E(Y^2) - \left(E(Y)\right)^2 = 1, \text{ so}$$ $$E(X^2) = \operatorname{Var}(X) + \left(E(X)\right)^2 = 8 \text{ and } E(Y^2) = \operatorname{Var}(Y) + \left(E(Y)\right)^2 = 50.$$

$E(X)E(Y) = E(XY)$, as $X$ and $Y$ are independent.

$$\begin{array}{l} \operatorname{Var}(5X + 2Y - 2) = E((5X + 2Y - 2)^2) - \left(E(5X + 2Y - 2)\right)^2 = E((5X + 2Y - 2)^2) - 484 = \\ = E(25X^2 + 4Y^2 + 4 + 20XY - 20X - 8Y) - 484 = \\ = E(25X^2) + E(4Y^2) + E(4) + E(20XY) + E(-20X) + E(-8Y) - 484 = \\ = 25E(X^2) + 4E(Y^2) + E(4) + 20E(X)E(Y) - 20E(X) - 8E(Y) - 484 = 104 \\ \end{array}$$

by the definition of the variance.

Other method uses properties of variances:

$$\begin{array}{l} \operatorname{Var}(5X + 2Y - 2) = \operatorname{Var}(5X + 2Y) = |X \text{ and } Y \text{ are independent variables}| = \\ = \operatorname{Var}(5X) + \operatorname{Var}(2Y) = 5^2\operatorname{Var}(X) + 2^2\operatorname{Var}(Y) = 25\operatorname{Var}(X) + 4\operatorname{Var}(Y) = \\ = 25 \cdot 4 + 4 \cdot 1 = 104 \\ \end{array}$$

Answer:

1) $E(3X + 7) = 13$, $\operatorname{Var}(3X + 7) = 36$.

2) $E(5X + 2Y - 2) = 22$, $\operatorname{Var}(5X + 2Y - 2) = 104$.

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