# Math Help - Binomial distribution problem

1. ## Binomial distribution problem

Show that the variance of a binomial distribution with parameters n and p cannot exceed $\frac{n}{4}$

2. Originally Posted by zorro
Show that the variance of a binomial distribution with parameters n and p cannot exceed $\frac{n}{4}$
The graph of Var(X) = np(1-p) versus p is a parabola over the domain $0 \leq p \leq 1$. Do you know how to find the maximum turning point of a parabola ....?

3. You need to show that pq never exceeds 1/4.

Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

You should get p=1/2 produces the maximum.

4. ## no

Originally Posted by mr fantastic
The graph of Var(X) = np(1-p) versus p is a parabola over the domain $0 \leq p \leq 1$. Do you know how to find the maximum turning point of a parabola ....?
No

5. ## difficulty understanding

Originally Posted by skamoni
You need to show that pq never exceeds 1/4.

Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

You should get p=1/2 produces the maximum.

From where did q come in to this question

6. Differentiate the given Parabola (In this case -p^2 + p ) and put it equal to zero, to obtain the value of p for which the parabola is maximised.

I let q=1-p (I think this is standard notation).

pq=p(1-p) = p - p^2

7. Originally Posted by mr fantastic
The graph of Var(X) = np(1-p) versus p is a parabola over the domain $0 \leq p \leq 1$. Do you know how to find the maximum turning point of a parabola ....?
Originally Posted by zorro
No
The turning point lies between the two intercepts. The intercepts are p = 0 and p = 1. So the (maximum) turning point occurs at p = 1/2. Substitute = 1/2 into Var(X) = np(1-p) to get the required result.

8. Originally Posted by skamoni
You need to show that pq never exceeds 1/4.

Show that pq = -p^2 + p and find the the stationary point of this parabola (Which is a maximum).

You should get p=1/2 produces the maximum.
Originally Posted by zorro
From where did q come in to this question
q= 1 - p. It's a very standard notation. The variance of a binomial distribution is often written npq.

9. Originally Posted by zorro
From where did q come in to this question
If you dont know Analytical geometry, there are other ways to see it as well:

You can try:

1) Completing the squares.

$np(1-p) = n(p - p^2) = n\left(\frac14 - \left( p - \frac12\right)^2\right)$. But $\left( p - \frac12\right)^2 \geq 0$, so what can you conclude?

2) AM-GM inequality,

You can apply it(since p and q are non-negative), to get $\frac{p+q}2 \geq \sqrt{pq} \Rightarrow \sqrt{pq} \leq \frac12 \Rightarrow npq \leq \frac{n}4$