185k views
5 votes
P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

1 Answer

4 votes
Sounds like you're asked to find
a such that


\displaystyle\sum_(k=2)^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1

In other words, find
a that satisfies


a^2+a^3+a^4=1

We can factorize this as


a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0

In order that
\mathbb P(k) describes a probability distribution, require that
\mathbb P(k)\ge0 for all
k, which means we can ignore the possibility of
a=-1.

Let
a=y+\frac xy.


a^3+a-1=\left(y+\frac xy\right)^3+\left(y+\frac xy\right)-1=0

\left(y^3+3xy+\frac{3x^2}y+(x^3)/(y^3)\right)+\left(y+\frac xy\right)-1=0

Multiply both sides by
y^3.


y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0

We want to find
x\\eq0 that removes the quartic and quadratic terms from the equation, i.e.


\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\frac13

so the cubic above transforms to


y^6-y^3-\frac1{27}=0

Substitute
y^3=z and we get


z^2-z-\frac1{27}=0\implies z=(9+√(93))/(18)

\implies y=\sqrt[3]{(9+√(93))/(18)}

\implies a=\sqrt[3]{(9+√(93))/(18)}-\frac13\sqrt[3]{(18)/(9+√(93))}
User Emperor Eto
by
8.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.