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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).
Solution 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
∑xi P(xi)0 and ∑xi2 ·P(xi)= 1
E(X) = xi P(xi) = 0
Var (X) = ∑xi2 · P(xi) - [E(X)]2
= 1 - 0 = 1
Hence, E(X) = 0, Var (X) = 1.
Solution 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
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