Determine-the-following-sum-in-terms-of-n-S-n-i-1-n-i-2-q-i-1-where-0-lt-q-lt-1-For-the-distribution-of-the-discrete-random-variable-X-given-as-

Question Number 683 by 112358 last updated on 24/Feb/15

D e t e r m i n e t h e f o l l o w i n g s u m i n

t e r m s o f n .

S_{n} = \underset{i = 1}{\sum^{n}} i^{2} q^{i - 1}

w h e r e 0 < q < 1 .

F o r t h e d i s t r i b u t i o n o f t h e d i s c r e t e

r a n d o m v a r i a b l e X g i v e n a s

X ∽ G e o (p)

p r o v e t h a t

V a r (X) = \frac{q}{p^{2}}

w h e r e q = 1 - p, i f

V a r (X) = Σ {x^{2} P (X = x)} - E^{2} (X)

f o r a g i v e n d i s c r e t e r a n d o m

v a r i a b l e X .

Commented by prakash jain last updated on 24/Feb/15

S_{n} = 1^{2} + 2^{2} q + 3^{2} q^{2} + 4^{2} q^{3} + \dots + n^{2} q^{n - 1} \dots . (i)

{q S}_{n} = 1^{2} q + 2^{2} q^{2} + 3^{2} q^{2} + \dots + {(n - 1)}^{2} q^{n - 1} + n^{2} q^{n} \dots (ii)

Subtracting (ii) from (i)

(1 - q) S_{n} = 1 + 3 q + 5 q^{2} + \dots . . + (2 n - 1) q^{n - 1} - n^{2} q^{n} \dots (iii)

q (1 - q) S_{n} = q + 3 q^{2} + \dots . . + (2 n - 3) q^{n - 1} + (2 n - 1) q^{n} - n^{2} q^{n + 1} \dots (iv)

Subtracting (iv) from (iii)

{(1 - q)}^{2} S_{n} = 1 + 2 q + 2 q^{2} + . . + 2 q^{n - 1} - (n^{2} - 2 n + 1) q^{n} + n^{2} q^{n + 1}

{(1 - q)}^{2} S_{n} = 1 + 2 \frac{q - q^{n}}{1 - q} - {(n - 1)}^{2} q^{n} + n^{2} q^{n + 1}

S_{n} = \frac{1}{{(1 - q)}^{2}} [1 + 2 \frac{q - q^{n}}{1 - q} - {(n - 1)}^{2} q^{n} + n^{2} q^{n + 1}]

I {don}^{'} t think you can calculate S_{n} only in terms

of n (you will also need q)

Commented by 112358 last updated on 24/Feb/15

W o u l d i t b e s a f e t o a s s u m e t h a t q

i s a c o n s t a n t s u c h t h a t q \in (0, 1) ?

Commented by prakash jain last updated on 24/Feb/15

For a give distribution as mention in problem

q = 1 - p so q is independent of i and n . So it is a

constant for the summation .

Answered by prakash jain last updated on 24/Feb/15

Expected Value

E (x) = \underset{i = 1}{\sum^{\infty}} i . {(1 - p)}^{i - 1} p

E = p + 2 (1 - p) p + 3 {(1 - p)}^{2} p + \dots

(1 - p) E = + (1 - p) p + 2 {(1 - p)}^{2} p + \dots

p E = p + (1 - p) p + {(1 - p)}^{2} p + . .

p E = \frac{p}{1 - (1 - p)} = 1 \Rightarrow E (x) = \frac{1}{p}

Variance

Var (x) = \underset{i = 1}{\sum^{\infty}} i^{2} {(1 - p)}^{i - 1} p - {[E (x)]}^{2} (i)

S = \underset{i = 1}{\sum^{\infty}} i^{2} {(1 - p)}^{i - 1} p

S = p + 2^{2} (1 - p) p + 3^{2} {(1 - p)}^{2} p + . .

(1 - p) S = (1 - p) p + 2^{2} {(1 - p)}^{2} p + \dots

S ubtracting

p S = p + 3 (1 - p) p + 5 {(1 - p)}^{2} p + \dots

S = 1 + 3 (1 - p) + 5 {(1 - p)}^{2} + \dots

(1 - p) S = (1 - p) + 3 {(1 - p)}^{2} + 5 {(1 - p)}^{3} + . .

S ubtracting

p S = 1 + 2 (1 - p) + 2 {(1 - p)}^{2} + \dots

p S = 1 + \frac{2 (1 - p)}{1 - (1 - p)} = 1 + \frac{2 - 2 p}{p} = \frac{2}{p} - 1

S = \frac{2}{p^{2}} - \frac{1}{p}

Substituting S in (i)

Var (x) = \frac{2}{p^{2}} - \frac{1}{p} - {[E (x)]}^{2} = \frac{2}{p^{2}} - \frac{1}{p} - \frac{1}{p^{2}}

= \frac{1}{p^{2}} - \frac{1}{p} = \frac{1 - p}{p^{2}} = \frac{q}{p^{2}}