Question

given below is a bivariate distribution for the random variables x and y.
f(x,y) x y
0.3 60 90
0.6 30 30
0.1 50 60
a. compute the expected value and the variance for x and y.
e(x) =
e(y) =
var(x) =
var(y) =
b. develop a probability distribution for x + y (to 2 decimals).
x + y f(x + y)
150
60
110
c. using the result of part (b), compute e(x + y) and var(x + y).
e(x + y) =

Explanation:

Step1: Recall expected - value formula for discrete random variables

The expected value of a discrete random variable $X$ is $E(X)=\sum_{i}x_ip_i$. For $x$:
$E(x)=0.3\times60 + 0.6\times30+0.1\times50=18 + 18+5=41$
For $y$:
$E(y)=0.3\times90 + 0.6\times30+0.1\times60=27 + 18+6=51$

Step2: Recall variance formula for discrete random variables

The variance of a discrete random variable $X$ is $Var(X)=\sum_{i}(x_i - E(X))^{2}p_i$.
For $x$:
$(60 - 41)^{2}\times0.3+(30 - 41)^{2}\times0.6+(50 - 41)^{2}\times0.1$
$=19^{2}\times0.3+(- 11)^{2}\times0.6 + 9^{2}\times0.1$
$=361\times0.3+121\times0.6 + 81\times0.1$
$=108.3+72.6+8.1 = 189$
For $y$:
$(90 - 51)^{2}\times0.3+(30 - 51)^{2}\times0.6+(60 - 51)^{2}\times0.1$
$=39^{2}\times0.3+(-21)^{2}\times0.6 + 9^{2}\times0.1$
$=1521\times0.3 + 441\times0.6+81\times0.1$
$=456.3+264.6 + 8.1=729$

Step3: Find probability distribution of $x + y$

When $x = 60,y = 90,x + y=150$ with probability $0.3$
When $x = 30,y = 30,x + y=60$ with probability $0.6$
When $x = 50,y = 60,x + y=110$ with probability $0.1$
So the probability distribution of $x + y$ is:

$x + y$	$f(x + y)$
$60$	$0.60$
$110$	$0.10$

Step4: Compute $E(x + y)$

$E(x + y)=150\times0.3+60\times0.6 + 110\times0.1=45+36 + 11=92$

Step5: Compute $Var(x + y)$

$(150 - 92)^{2}\times0.3+(60 - 92)^{2}\times0.6+(110 - 92)^{2}\times0.1$
$=58^{2}\times0.3+(-32)^{2}\times0.6+18^{2}\times0.1$
$=3364\times0.3 + 1024\times0.6+324\times0.1$
$=1009.2+614.4+32.4 = 1656$

Answer:

$E(x)=41$
$E(y)=51$
$Var(x)=189$
$Var(y)=729$

$x + y$	$f(x + y)$
$60$	$0.60$
$110$	$0.10$

$E(x + y)=92$
$Var(x + y)=1656$