Show that p (y) p (y - 1) = (n - y + 1) p yq > 1 if y p (y - 1) if y is small (y < (n + 1) p) and p (y) (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on. (n + 1) p> y

Question is not well presented

Consider the binomial distribution with n trials and P (S) = p.

Show that

p (y) / p (y - 1) = (n - y + 1) p / yq > 1 if y < (n + 1) p.

This establishes that p (y) > p (y - 1) if y is small (y < (n + 1) p) and p (y) (n + 1) p). Thus, successive binomial probabilities increase for a while and decrease from then on.

Answer:

See Explanation Below

Step-by-step explanation:

Given that

p (y) / p (y - 1) = (n - y + 1) p / yq > 1 if y < (n + 1) p.

First, we make the following assumption

p (y) / p (y - 1) = (n - y + 1) p / yq = 1;

So, we have

(n - y + 1) p / yq = 1

(n - y + 1) p = yq

Note that p + q = 1;.

So, q = 1 - p

Substitute 1-p for q in the above expression

(n - y + 1) p = y (1-p)

np - py + p = y - py

Solve for y

... Collect like terms

np + p = y - py + py

np + p = y;

So, y = np + p

y = (n + 1) p

From the above,

We know that

p (y) / p (y - 1) = 1

if y = (n + 1) p.

Similarly, we've also obtain that

p (y) / p (y - 1) > 1 if y < (n + 1) p.