Question

Solve the following problem :

Let the p. m. f. of the r. v. X be

`"P"(x) = {((3 - x)/(10)", ","for"  x = -1", "0", "1", "2.),(0,"otherwise".):}`
Calculate E(X) and Var(X).

Answer 1

P(X)=`(3-x) /10`

X takes values -1, 0, 1, 2

P(X = -1)= P(-1) = `(3+1) /10 =4/ 10`

P(X = 0)= P(0) = `(3-0) /10 =3/ 10`

P(X = 1)= P(1) = `(3-1) /10 =2/ 10`

P(X = 2)= P(2) = `(3-2) /10 =1/ 10`

We construct the following table to calculate the mean and variance of X :

xi	P (xi)	xi P (xi)	xi2 P (xi)
-1	`4/10`	-`4/10`	`4/10`
0	`3/10`	0	0
1	`2/ 10`	`2/10`	`2/10`
2	`1/10`	`2/10`	`4/10`
Total	1	0	1

From the table

∑x_i P(x_i)0 and ∑x_i² ·P(x_i)= 1

E(X) = x_i P(x_i) = 0

Var (X) = ∑x_i² · P(x_i) - [E(X)]²

= 1 - 0 = 1

Hence, E(X) = 0, Var (X) = 1.

Answer 2

E(X) = `sum_("i" = 1)^4x_"i""P"(x_"i")`

= `-1 xx ((3 - (-1))/10) + 0 xx ((3 - 0)/10) + 1 xx ((3 - 1)/10) + 2 xx ((3 - 2)/10)`

= `(-4 + 0 + 2 + 2)/10`

= 0

E(X²) = `sum_("i" = 1)^4x_"i"^2"P"(x_"i")`

= `(-1)^2 xx 4/10 + (0)^2 xx 3/10 + (1)^2 xx 2/10 + (2)^2 xx 1/10`

= `(4 + 0 + 2 + 4)/10`

= 1

Var(X) = E(X²) – [E(X)]²

= 1 – (0)²

= 1