Question Number 683 by 112358 last updated on 24/Feb/15
$${Determine}\:{the}\:{following}\:{sum}\:{in} \\ $$$${terms}\:{of}\:{n}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{S}_{{n}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{i}^{\mathrm{2}} {q}^{{i}−\mathrm{1}} \\ $$$${where}\:\mathrm{0}\:<\:{q}\:<\:\mathrm{1}. \\ $$$${For}\:{the}\:{distribution}\:{of}\:{the}\:{discrete} \\ $$$${random}\:{variable}\:{X}\:{given}\:{as} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{X}\backsim{Geo}\left({p}\right) \\ $$$${prove}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Var}\left({X}\right)=\frac{{q}}{{p}^{\mathrm{2}} } \\ $$$${where}\:{q}=\mathrm{1}−{p},\:{if}\: \\ $$$${Var}\left({X}\right)=\Sigma\left\{{x}^{\mathrm{2}} {P}\left({X}={x}\right)\right\}−{E}^{\mathrm{2}} \left({X}\right) \\ $$$${for}\:{a}\:{given}\:{discrete}\:{random}\: \\ $$$${variable}\:{X}. \\ $$
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 +... +n^2 q^(n−1) ....(i) qS_n = 1^2 q +2^2 q^2 +3^2 q^2 +...+(n−1)^2 q^(n−1) +n^2 q^n ...(ii) Subtracting (ii) from (i) (1−q)S_n =1+3q+5q^2 +.....+(2n−1)q^(n−1) −n^2 q^n ...(iii) q(1−q)S_n = q+3q^2 +.....+(2n−3)q^(n−1) +(2n−1)q^n −n^2 q^(n+1) ...(iv) Subtracting (iv) from (iii) (1−q)^2 S_n =1+2q+2q^2 +..+2q^(n−1) −(n^2 −2n+1)q^n +n^2 q^(n+1) (1−q)^2 S_n =1+2 ((q−q^n )/(1−q))−(n−1)^2 q^n +n^2 q^(n+1) S_n =(1/((1−q)^2 ))[1+2((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)](https://www.tinkutara.com/question/Q684.png)
$${S}_{\mathrm{n}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} {q}+\mathrm{3}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} {q}^{\mathrm{3}} +…\:+{n}^{\mathrm{2}} {q}^{{n}−\mathrm{1}} \:\:\:\:\:\:\:\:….\left(\mathrm{i}\right) \\ $$$${qS}_{\mathrm{n}} =\:\:\:\:\:\mathrm{1}^{\mathrm{2}} {q}\:\:\:+\mathrm{2}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} {q}^{\mathrm{2}} +…+\left({n}−\mathrm{1}\right)^{\mathrm{2}} {q}^{{n}−\mathrm{1}} +{n}^{\mathrm{2}} {q}^{{n}} \:\:\:\:…\left(\mathrm{ii}\right) \\ $$$$\mathrm{Subtracting}\:\left(\mathrm{ii}\right)\:\mathrm{from}\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{1}−{q}\right){S}_{{n}} =\mathrm{1}+\mathrm{3}{q}+\mathrm{5}{q}^{\mathrm{2}} +…..+\left(\mathrm{2}{n}−\mathrm{1}\right){q}^{{n}−\mathrm{1}} −{n}^{\mathrm{2}} {q}^{{n}} \:\:\:\:…\left(\mathrm{iii}\right) \\ $$$${q}\left(\mathrm{1}−{q}\right){S}_{{n}} =\:\:\:\:\:\:\:\:\:{q}+\mathrm{3}{q}^{\mathrm{2}} +…..+\left(\mathrm{2}{n}−\mathrm{3}\right){q}^{{n}−\mathrm{1}} +\left(\mathrm{2}{n}−\mathrm{1}\right){q}^{{n}} −{n}^{\mathrm{2}} {q}^{{n}+\mathrm{1}} …\left(\mathrm{iv}\right) \\ $$$$\mathrm{Subtracting}\:\left(\mathrm{iv}\right)\:\mathrm{from}\:\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {S}_{{n}} =\mathrm{1}+\mathrm{2}{q}+\mathrm{2}{q}^{\mathrm{2}} +..+\mathrm{2}{q}^{{n}−\mathrm{1}} −\left({n}^{\mathrm{2}} −\mathrm{2}{n}+\mathrm{1}\right){q}^{{n}} +{n}^{\mathrm{2}} {q}^{{n}+\mathrm{1}} \\ $$$$\left(\mathrm{1}−{q}\right)^{\mathrm{2}} {S}_{{n}} =\mathrm{1}+\mathrm{2}\:\:\frac{{q}−{q}^{{n}} }{\mathrm{1}−{q}}−\left({n}−\mathrm{1}\right)^{\mathrm{2}} {q}^{{n}} +{n}^{\mathrm{2}} {q}^{{n}+\mathrm{1}} \\ $$$${S}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{1}−{q}\right)^{\mathrm{2}} }\left[\mathrm{1}+\mathrm{2}\frac{{q}−{q}^{{n}} }{\mathrm{1}−{q}}−\left({n}−\mathrm{1}\right)^{\mathrm{2}} {q}^{{n}} +{n}^{\mathrm{2}} {q}^{{n}+\mathrm{1}} \right] \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{calculate}\:{S}_{{n}} \:\mathrm{only}\:\mathrm{in}\:\mathrm{terms} \\ $$$$\mathrm{of}\:{n}\:\left(\mathrm{you}\:\mathrm{will}\:\mathrm{also}\:\mathrm{need}\:{q}\right) \\ $$
Commented by 112358 last updated on 24/Feb/15
$${Would}\:{it}\:{be}\:{safe}\:{to}\:{assume}\:{that}\:{q} \\ $$$${is}\:{a}\:{constant}\:{such}\:{that}\:{q}\in\left(\mathrm{0},\mathrm{1}\right)? \\ $$
Commented by prakash jain last updated on 24/Feb/15
$$\mathrm{For}\:\mathrm{a}\:\mathrm{give}\:\mathrm{distribution}\:\mathrm{as}\:\mathrm{mention}\:\mathrm{in}\:\mathrm{problem} \\ $$$${q}=\mathrm{1}−{p}\:\:\mathrm{so}\:{q}\:\mathrm{is}\:\mathrm{independent}\:\mathrm{of}\:{i}\:\mathrm{and}\:{n}.\:\mathrm{So}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{for}\:\mathrm{the}\:\mathrm{summation}. \\ $$
Answered by prakash jain last updated on 24/Feb/15
![Expected Value E(x)=Σ_(i=1) ^∞ i.(1−p)^(i−1) p E= p+2(1−p)p+3(1−p)^2 p+... (1−p)E= +(1−p)p+2(1−p)^2 p+... pE =p+(1−p)p+(1−p)^2 p+.. pE=(p/(1−(1−p)))=1⇒E(x)=(1/p) Variance Var(x)=Σ_(i=1) ^∞ i^2 (1−p)^(i−1) p−[E(x)]^2 (i) S=Σ_(i=1) ^∞ 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+... Subtracting pS=p+3(1−p)p+5(1−p)^2 p+... S=1+3(1−p)+5(1−p)^2 +... (1−p)S=(1−p)+3(1−p)^2 +5(1−p)^3 +.. Subtracting pS=1+2(1−p)+2(1−p)^2 +... pS=1+((2(1−p))/(1−(1−p)))=1+((2−2p)/p)=(2/p)−1 S=(2/p^2 ) −(1/p) Substituting S in (i) Var(x)=(2/p^2 ) −(1/p)−[E(x)]^2 =(2/p^2 ) −(1/p) −(1/p^2 ) =(1/p^2 )−(1/p)=((1−p)/p^2 )=(q/p^2 )](https://www.tinkutara.com/question/Q688.png)
$$\mathrm{Expected}\:\mathrm{Value} \\ $$$${E}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}.\left(\mathrm{1}−{p}\right)^{{i}−\mathrm{1}} {p} \\ $$$${E}=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}+\mathrm{2}\left(\mathrm{1}−{p}\right){p}+\mathrm{3}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+… \\ $$$$\left(\mathrm{1}−{p}\right){E}=\:\:\:\:\:+\left(\mathrm{1}−{p}\right){p}+\mathrm{2}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+… \\ $$$${pE}\:={p}+\left(\mathrm{1}−{p}\right){p}+\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+.. \\ $$$${pE}=\frac{{p}}{\mathrm{1}−\left(\mathrm{1}−{p}\right)}=\mathrm{1}\Rightarrow{E}\left({x}\right)=\frac{\mathrm{1}}{{p}} \\ $$$$\mathrm{Variance} \\ $$$$\mathrm{Var}\left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{{i}−\mathrm{1}} {p}−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \:\:\:\:\:\:\:\:\left(\mathrm{i}\right) \\ $$$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{i}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{{i}−\mathrm{1}} {p} \\ $$$${S}={p}+\mathrm{2}^{\mathrm{2}} \left(\mathrm{1}−{p}\right){p}+\mathrm{3}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+.. \\ $$$$\left(\mathrm{1}−{p}\right){S}=\left(\mathrm{1}−{p}\right){p}+\mathrm{2}^{\mathrm{2}} \left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+… \\ $$$${S}\mathrm{ubtracting} \\ $$$${pS}={p}+\mathrm{3}\left(\mathrm{1}−{p}\right){p}+\mathrm{5}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} {p}+… \\ $$$${S}=\mathrm{1}+\mathrm{3}\left(\mathrm{1}−{p}\right)+\mathrm{5}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} +… \\ $$$$\left(\mathrm{1}−{p}\right){S}=\left(\mathrm{1}−{p}\right)+\mathrm{3}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} +\mathrm{5}\left(\mathrm{1}−{p}\right)^{\mathrm{3}} +.. \\ $$$${S}\mathrm{ubtracting} \\ $$$${pS}=\mathrm{1}+\mathrm{2}\left(\mathrm{1}−{p}\right)+\mathrm{2}\left(\mathrm{1}−{p}\right)^{\mathrm{2}} +… \\ $$$${pS}=\mathrm{1}+\frac{\mathrm{2}\left(\mathrm{1}−{p}\right)}{\mathrm{1}−\left(\mathrm{1}−{p}\right)}=\mathrm{1}+\frac{\mathrm{2}−\mathrm{2}{p}}{{p}}=\frac{\mathrm{2}}{{p}}−\mathrm{1} \\ $$$${S}=\frac{\mathrm{2}}{{p}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{p}} \\ $$$$\mathrm{Substituting}\:{S}\:\mathrm{in}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{Var}\left({x}\right)=\frac{\mathrm{2}}{{p}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{p}}−\left[{E}\left({x}\right)\right]^{\mathrm{2}} =\frac{\mathrm{2}}{{p}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{p}}\:−\frac{\mathrm{1}}{{p}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{p}}=\frac{\mathrm{1}−{p}}{{p}^{\mathrm{2}} }=\frac{{q}}{{p}^{\mathrm{2}} } \\ $$