Question

Random variables X and Y have following distributions. P(X = -1) = 3/4, P(X = 3) = 1/4 P(Y = -3) = 1/2, P(Y = 2) = 1/2 a) Using the moment generating functions for random variables above find: E[X+Y) b) Using the moment generating functions for random variables above find: Var(X+Y)

Answer 1

GIVEN :

Distribution of given random variable X:

random variable X	-1	3
Probability	3/4	1/4

Distribution of given random variable Y:

random variable X	-3	2
Probability	1/2	1/2

WE KNOW THAT:

a)

E(X) = ΣτΡ()

So, E[X] = (-1*3/4) + (3*1/4)

= -3/4 + 3/4 = 0

So, E[Y] = (-3*1/2) + (2*1/2)

= -3/2 + 2/2 = -1/2

E[X] = 0, And

E[Y] = -1/2

Also,

E(X+Y) = E(X) + E(Y)

so , from these moment generating functions

E[X+Y] = 0 + (-1/2) = -1/2

E[X+Y] = -1/2

b)

we know that

${\displaystyle \operatorname {Var} (X)=\left(\sum _{i=1}^{n}p_{i}x_{i}^{2}\right)-\mu ^{2},}$

where  $\mu$ is the expected value.

${\displaystyle \mu =\sum _{i=1}^{n}p_{i}x_{i}.}$

E(X) = M = 0

Var(X) = ((-1)2 * 3/4) + ((3)2 * 1/4) - 02 = 3

Also,

$E(Y) = \mu = -1/2$

Also,

Var(X + Y) = Var(X - Y) = Var(X) + Var(Y)

Var(X+Y) = 3 +9/4

Var(X+Y) = 5.25

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