Find the expected value, variance and standard deviation of the random variable whose p.m.f.’s are given below :

x = x | 1 | 2 | 3 | ... | n |

P (X = x) | `1/n` | `1/n` | `1/n` | ... | `1/n` |

#### Solution

We construct the following table to find the expected value, variance, and standard deviation:

x_{i} |
P (x_{i}) |
x_{i}·P (x_{i}) |
x_{i }^{2}·P (x_{i}) = x_{i }× x_{i}·P (x_{i}) |

1 | `1/n` | `1/n` | `1/n` |

2 | `1/n` | `2/n` | `2^2/n` |

3 | `1/n` | `3/n` | `3^2/n` |

... | ... | ... | ... |

n | `1/n` | `n/n` | `n^2/n` |

From the table,

∑x_{i} . P(x_{i}) = `1/n + 2/n + 3/n + ... + n/n `

= `1/n ( 1 + 2 + 3 + ... + n)`

= `1/n sum _( r=1)^n r = 1/n xx (n(n+1 ))/2`

= `(n+1)/2`

∑x_{i2} . P(x_{i}) = `1/n + 2^2/n + 3^2/n + ... + n^2/n `

= `1/n ( 1^2 + 2^2 + 3^2 + ... + n^2)`

= `1/n sum _( r=1)^n r^2 = 1/n xx (n(n+1 )(2n +1))/6`

= `((n+1)(2n +1))/6``

∴ expected value = E(X) = ∑ x_{i }· P(x_{i})

= `(n+1)/2`

Variance = V(X) = ∑ x_{i}^{2} . P (x_{i}) - [ ∑ x_{i}·P (x_{i}) ]^{2}

= `((n+1)(2n + 1))/6 - ((n + 1)/2)^2`

= `(n+1)/2 [(2n + 1)/ 3 -(n + 1) /2 ]`

= `(n+1)/2 [(4n+2-3n-3)/6 ]`

=`((n + 1)(n - 1))/12`

=` (n^2 - 1)/12`

Standard deviation = `sqrt (V(X)`

= `sqrt ((n^2-1)/12)`

= `(sqrt (n^2-1))/(2sqrt 3)`