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Question Number 192105 by Spillover last updated on 08/May/23
Answered by mehdee42 last updated on 09/May/23
a) lema : if 0<x<1 ⇒ Σ_(i=0) ^∞ ix^i =(x/((1−x)^2 )) & Σ_(i=0) ^∞ i^2 x^i =((x(1+x))/((1−x)^3 )) thus E(x)=Σ xf(x)=(2/3)Σ_(i=0) ^∞ i((1/3))^i =(1/2) ✓ b) Var(x)=E(x^2 )−E^2 (x) E(x^2 )=Σx^2 f(x)=(2/3)Σ_(i=0) ^∞ i^2 ((1/3))^i =(2/3)×(4/9)×((27)/8) =1 ⇒Var(x)=1−(1/4)=(3/4) ✓ c) P(0)+P(1)+P(2)+P(3)= (2/3)(0+(1/3)+(1/9)+(1/(27)))=((26)/(81)) ✓
$$\left.\:{a}\right)\:\:\:{lema}\::\:{if}\:\:\mathrm{0}<{x}<\mathrm{1}\:\Rightarrow\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{ix}^{{i}} =\frac{{x}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\:\:\&\:\:\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{i}^{\mathrm{2}} {x}^{{i}} =\frac{{x}\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} }\: \\ $$$${thus}\:\:\:{E}\left({x}\right)=\Sigma\:{xf}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\:{i}\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{i}} =\frac{\mathrm{1}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$$$\left.{b}\right)\:{Var}\left({x}\right)={E}\left({x}^{\mathrm{2}} \right)−{E}^{\mathrm{2}} \left({x}\right) \\ $$$${E}\left({x}^{\mathrm{2}} \right)=\Sigma{x}^{\mathrm{2}} {f}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{i}^{\mathrm{2}} \:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{i}} =\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{27}}{\mathrm{8}}\:=\mathrm{1} \\ $$$$\Rightarrow{Var}\left({x}\right)=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{4}}\:\checkmark \\ $$$$\left.{c}\right)\:{P}\left(\mathrm{0}\right)+{P}\left(\mathrm{1}\right)+{P}\left(\mathrm{2}\right)+{P}\left(\mathrm{3}\right)=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{0}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{27}}\right)=\frac{\mathrm{26}}{\mathrm{81}}\:\checkmark \\ $$
Commented by Spillover last updated on 12/May/23
thanks
$${thanks} \\ $$
Answered by Spillover last updated on 12/May/23
From factorial generating function ∅_x (t)=E(t^x ) E(t^x )=Σ_x ^∞ (2/3)t^x ((1/3))^x (2/3)Σ_(x=0) ^∞ t^x ((1/3))^x = (2/3)Σ_(x=0) ^∞ ((t/3))^x (2/3)Σ_(x=0) ^∞ ((t/3))^x =(2/3)(1+(t/3)+(t^2 /3^(2 ) )+(t^3 /3^3 )+...) let y=(t/3) E(t^x )=(2/3)(1+y+y^2 +y^3 +...) from s_∞ =(G_1 /(1−r)) s_∞ =(1/(1−r)) but r=y=(t/3) s_∞ =(1/(1−r)) =(1/(1−(t/3)))=(3/(3−t)) E(t^x )=(2/3)(1+y+y^2 +y^3 +...) E(t^x )=(2/3)((3/(3−t)))=(2/(3−t)) ∅_x (t)=(2/(3−t)) ∅_x ^′ (t)=(2/((3−t)^2 )) ∅_x ^(′′) (t)=(2/((3−t)^3 )) ∅_x ^(′′′) (t)=(2/((3−t)^3 )) E(x)=∅_x ^′ (t)=(2/((3−t)^2 )) t=1 E(x)=((2/((3−1)^2 ))) =(1/2) var(x)=E(x)−[E(x)]^2 var(x)= ∅_x ^(′′) (t)+E(x)−[E(x)]^2 var(x)= (1/2)+(1/2)−((1/2))^2 =(3/4)
$${From}\:{factorial}\:{generating}\:{function}\: \\ $$$$\varnothing_{{x}} \left({t}\right)={E}\left({t}^{{x}} \right) \\ $$$${E}\left({t}^{{x}} \right)=\sum_{{x}} ^{\infty} \frac{\mathrm{2}}{\mathrm{3}}{t}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{x}} \:\:\:\:\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}{t}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{x}} =\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{t}}{\mathrm{3}}\right)^{{x}} \\ $$$$\frac{\mathrm{2}}{\mathrm{3}}\underset{{x}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{t}}{\mathrm{3}}\right)^{{x}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+\frac{{t}}{\mathrm{3}}+\frac{{t}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}\:} }+\frac{{t}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }+…\right) \\ $$$${let}\:{y}=\frac{{t}}{\mathrm{3}} \\ $$$${E}\left({t}^{{x}} \right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{y}+{y}^{\mathrm{2}} +{y}^{\mathrm{3}} +…\right) \\ $$$${from}\:{s}_{\infty} =\frac{{G}_{\mathrm{1}} }{\mathrm{1}−{r}}\:\:\:\:\:\:{s}_{\infty} =\frac{\mathrm{1}}{\mathrm{1}−{r}}\:\:\:\:{but}\:{r}={y}=\frac{{t}}{\mathrm{3}} \\ $$$${s}_{\infty} =\frac{\mathrm{1}}{\mathrm{1}−{r}}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{{t}}{\mathrm{3}}}=\frac{\mathrm{3}}{\mathrm{3}−{t}} \\ $$$${E}\left({t}^{{x}} \right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}+{y}+{y}^{\mathrm{2}} +{y}^{\mathrm{3}} +…\right)\:\:\:\:\:\:\:\:{E}\left({t}^{{x}} \right)=\frac{\mathrm{2}}{\mathrm{3}}\left(\frac{\mathrm{3}}{\mathrm{3}−{t}}\right)=\frac{\mathrm{2}}{\mathrm{3}−{t}} \\ $$$$\varnothing_{{x}} \left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}−{t}} \\ $$$$\varnothing_{{x}} ^{'} \left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{\mathrm{2}} }\:\:\:\:\:\varnothing_{{x}} ^{''} \left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{\mathrm{3}} }\:\:\:\:\:\varnothing_{{x}} ^{'''} \:\left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{\mathrm{3}} }\:\:\: \\ $$$${E}\left({x}\right)=\varnothing_{{x}} ^{'} \left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{\mathrm{2}} }\:\:\:\:\:\:{t}=\mathrm{1}\:\:\:{E}\left({x}\right)=\left(\frac{\mathrm{2}}{\left(\mathrm{3}−\mathrm{1}\right)^{\mathrm{2}} }\right)\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${var}\left({x}\right)={E}\left({x}\right)−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${var}\left({x}\right)=\:\varnothing_{{x}} ^{''} \left({t}\right)+{E}\left({x}\right)−\left[{E}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${var}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$
Answered by Spillover last updated on 12/May/23
From probability generating function of x G_x (t)=(2/(3−t)) G_x ^′ (t)=(2/((3−t)^(−2) )) G_x ^(′′) (t)=(2/((3−t)^(−3) )) G_x ′′′(t)=(2/((3−t)^(−4) )) G_x (t)=G_x (0)t^0 +G_x (0)t+G_x (0)(t^2 /(2!))++G_x (0)(t^3 /(3!))+... G_x (t)=(2/3)t^0 +(2/9)t+(4/(27))((t^2 /(2!)))+(2/(81))((t^3 /(3!)))+... G_x (t)=(2/3)t^0 +(2/9)t+(2/(27))t^2 +(2/(81))t^3 +... P(X=0)=(2/3) P(X=1)=(2/9) P(X=2)=(2/(27)) P(X=3)=(2/(81)) P(x)= P(X=0)+P(X=1)+ P(X=2)+ P(X=3) P(x)=(2/3)+(2/9)+(2/(27))=((36)/(81))
$${From}\:{probability}\:{generating}\:{function}\:{of}\:{x} \\ $$$${G}_{{x}} \left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}−{t}}\: \\ $$$${G}_{{x}} ^{'} \left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{−\mathrm{2}} }\:\:\:\:\:\:\:{G}_{{x}} ^{''} \left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{−\mathrm{3}} }\:\:\:\:\:\:{G}_{{x}} '''\left({t}\right)=\frac{\mathrm{2}}{\left(\mathrm{3}−{t}\right)^{−\mathrm{4}} }\: \\ $$$${G}_{{x}} \left({t}\right)={G}_{{x}} \left(\mathrm{0}\right){t}^{\mathrm{0}} +{G}_{{x}} \left(\mathrm{0}\right){t}+{G}_{{x}} \left(\mathrm{0}\right)\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}++{G}_{{x}} \left(\mathrm{0}\right)\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$${G}_{{x}} \left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{0}} +\frac{\mathrm{2}}{\mathrm{9}}{t}+\frac{\mathrm{4}}{\mathrm{27}}\left(\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}\right)+\frac{\mathrm{2}}{\mathrm{81}}\left(\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}\right)+… \\ $$$${G}_{{x}} \left({t}\right)=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{0}} +\frac{\mathrm{2}}{\mathrm{9}}{t}+\frac{\mathrm{2}}{\mathrm{27}}{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{81}}{t}^{\mathrm{3}} +… \\ $$$${P}\left({X}=\mathrm{0}\right)=\frac{\mathrm{2}}{\mathrm{3}}\:\:\:{P}\left({X}=\mathrm{1}\right)=\frac{\mathrm{2}}{\mathrm{9}}\:\:\:{P}\left({X}=\mathrm{2}\right)=\frac{\mathrm{2}}{\mathrm{27}}\:\:{P}\left({X}=\mathrm{3}\right)=\frac{\mathrm{2}}{\mathrm{81}} \\ $$$${P}\left({x}\right)=\:{P}\left({X}=\mathrm{0}\right)+{P}\left({X}=\mathrm{1}\right)+\:\:\:{P}\left({X}=\mathrm{2}\right)+\:\:{P}\left({X}=\mathrm{3}\right) \\ $$$${P}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{9}}+\frac{\mathrm{2}}{\mathrm{27}}=\frac{\mathrm{36}}{\mathrm{81}} \\ $$

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