Question

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

Answer 1

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))}`