Question

Problem 3 A discrete random variable Y takes values {k= 0, 1, 2, ...,} such that PLY Z k} = ()* for k 20. 1. Derive P[Y = k) for any k > 0. 2. Evaluate expectation, E[Y] = 3. Given E[Y(Y - 1)] = 15 , find variance of Y, Var[Y] =

Answer 1

Y is a Discrete R.V.

P(Y > K) *
; k = 0 , 1 , 2 , ............

1 . ) P ( Y = K )

P ( Y $\geq$ 0 ) = 1

P ( Y $\geq$ 1 ) =  $(\frac{3}{5})^^^1$

P ( Y$\geq$ 2 ) = $(\frac{3}{5})^^^2$

$\Rightarrow$ P ( Y = 0 ) =  P ( Y $\geq$ 0 ) -  P ( Y $\geq$ 1 ) = 1 - $(\frac{3}{5})^^^1$ =  $\frac{2}{5}$

P ( Y = 1 ) = P ( Y $\geq$ 1 ) - P ( Y $\geq$ 2 ) =  $(\frac{3}{5})^^^1$ -  $(\frac{3}{5})^^^2$ = $\frac{6}{25}$

Generalizing the above process :

Therefore , P ( Y = k ) = P ( Y $\geq$ k ) - P ( Y $\geq$ k + 1 )

=  $(\frac{3}{5})^^^k$ - $(\frac{3}{5})^^^k.(\frac{3}{5})$    ; k = 0 , 1 , 2 , .............

2 . ) E ( Y ) =  $\sum_{k=0}^\infty {k.P (Y=k)}$

=  $\sum_{k=0}^\infty {k.(\frac{3}{5})^^^k}$ - $\sum_{k=0}^\infty {k.(\frac{3}{5})^^^k . (\frac{3}{5})}$

=  $(1.(\frac{3}{5})+2.(\frac{3}{5})^2 + 3.(\frac{3}{5})^3 + ..............)$ - $\frac{3}{5}$  $(1.(\frac{3}{5})+2.(\frac{3}{5})^2 + 3.(\frac{3}{5})^3 + ..............)$

=  $\frac{2}{5}$  $(1.(\frac{3}{5})+2.(\frac{3}{5})^2 + 3.(\frac{3}{5})^3 + ..............)$

=  $\frac{6}{25}$$(1+2(\frac{3}{5})+3(\frac{3}{5})^2+............)$

This is an A.G.P Series with a = 1 , r = 0.6 , d =1 whose sum is given by :

$\frac{a}{1-r}+\frac{r.d}{(1-r)^2}$ ; where , a is first term , r is common ratio and d is the difference

$\Rightarrow$ E ( Y ) =  $\frac{6}{25}$  $(\frac{1}{1-\frac{3}{5}}+\frac{\frac{3}{5}.1}{(1-\frac{3}{5})^2})$ =   $\frac{3}{2}$ = 1.5

3 . ) Given : E ( Y ( Y - 1 ) ) = 15 / 2

$\Rightarrow$ E ( Y ² ) - E ( Y ) = 15 / 2

$\Rightarrow$ E ( Y ² ) =  $\frac{15}{2}+ E (Y )$ =  $\frac{15}{2}+\frac{3}{2}$ = 6

Since , V ( Y ) = E ( Y ² ) - [ E ( Y ) ] ²

$\Rightarrow$  V ( Y ) = 6 - 1.5 ²   = 6 - 2.25 =  $\frac{15}{4}$ = 3.75