# Thread: Check that each of the following is a valid probability function...

1. ## Check that each of the following is a valid probability function...

Check that each of the following is a valid probability function for a discrete random variable.

(i) g(n) = (2/3) (1/3)^(n-1), n=1,2,3,...

(ii) g(k) = (k+1) (2/3)^2 (1/3)^k, k=0,1,2,...

(Hint: use the formula (1-q)^(-2) = sum from zero to infinity of (n+1)q^n with 0<q<1.)

2. Originally Posted by feyomi
...for a discrete random variable.

(i) g(n) = (2/3) (1/3)^(n-1), n=1,2,3,...

(ii) g(k) = (k+1) (2/3)^2 (1/3)^k, k=0,1,2,...

(Hint: use the formula (1-q)^(-2) = sum from zero to infinity of (n+1)q^n with 0<q<1.)

hi feyomi,

if the formula represent probabilities, then the sum of all probabilities
must equal 1.

These are sums to infinity.

For (i)

$g(n)=\frac{2}{3}\ \left(\frac{1}{3}\right)^{n-1}$ n=1, 2, 3....

the sum of these "probabilities" is

$\frac{2}{3}\ \left(\frac{1}{3}\right)^0+\frac{2}{3}\ \left(\frac{1}{3}\right)^1+\frac{2}{3}\ \left(\frac{1}{3}\right)^2+....$

which is a geometric series, first term 2/3, common ratio 1/3, whose sum to infinity is $\frac{T_1}{1-r}=\frac{a}{1-r}$

$T1=a=\frac{2}{3},\ r=\frac{1}{3}$

$S_{\infty}=\frac{a}{1-r}=\frac{\frac{2}{3}}{1-\frac{1}{3}}=\frac{\left(\frac{2}{3}\right)}{\left (\frac{2}{3}\right)}=1$

All probabilities sum to 1, hence the function g(n) is valid.

For (ii), use the hint

$q=\frac{1}{3}$

Hence, the sum is

$\left(\frac{2}{3}\right)^2\left(1-\frac{1}{3}\right)^{-2}$

$=\left(\frac{2}{3}\right)^2\left(\frac{2}{3}\right )^{-2}$

As this is 1, this function is also valid.