Question

when calculating all the values of a Binomial Distribution the work and usually be simplified by first calculating b(0,n,theta) and then using the recursion formula b(x + 1 ;,n,theta)=(theta* (n-x)/(x+1)*(1-theta) ) *b(x;n,theta)
verify this formula and use it too calculate the values of the Binomial Distribution with n equal to 7 and theta is equal to 0.25

Answer 1

First
\[b(x; n, \theta) = \binom{n}{x} \theta^x (1-\theta)^{n-x}\]
Then,
\[
b(x+1; n, \theta) = \binom{n}{x+1} \theta^{x+1} (1-\theta)^{n-x-1}. \qquad (*)
\]
You want to show that (*) is equal to
\[
\left( \frac{\theta (n-x)}{(1-\theta)(x+1)} \right) \binom{n}{x} \theta^x (1-\theta)^{n-x}
\]
You have to look separately at
\[
\frac{n-x}{x+1}\binom{n}{x} \qquad\text{ and } \qquad \frac{\theta}{1-\theta} \theta^x (1-\theta)^{n-x}.
\]
Try this, and come back with what you obtained.