Maximizing b(n,k) in Bernoulli Trials Homework

[jgens]

Homework Statement

Suppose $0 < p < 1$. Define $b(n,k) = \binom{n}{k}p^k(1-p)^k$. For what value of $k$ is $b(n,k)$ a maximum?

Homework Equations

N/A

The Attempt at a Solution

Is there any way to get a nice closed form solution to this problem? I've already proved that it has a maximum so there must be some j such that $b(n,j)$ is maximal. Then we know $b(n,j+1) < b(n,j)$ and $b(n,j-1) < b(n,j)$. And I figure we can use these relations to figure out j or something like that. But I'm not sure if this approach will work.

Could someone help me with this please?

 

[Ray Vickson]

Homework Statement

Suppose $0 < p < 1$. Define $b(n,k) = \binom{n}{k}p^k(1-p)^k$. For what value of $k$ is $b(n,k)$ a maximum?

Homework Equations

N/A

The Attempt at a Solution

Is there any way to get a nice closed form solution to this problem? I've already proved that it has a maximum so there must be some j such that $b(n,j)$ is maximal. Then we know $b(n,j+1) < b(n,j)$ and $b(n,j-1) < b(n,j)$. And I figure we can use these relations to figure out j or something like that. But I'm not sure if this approach will work.

Could someone help me with this please?

Develop an expression for r(j) = b(n,j+1)/b(n,j) [which, by the way, is also useful for computing the b(n,k) recursively]. What can you say if r(j) > 1? If r(j) < 1?

RGV

 

[jgens]

Well, if I let $j$ be such that $b(n,j)$ is maximal, then

$$b(n,j-1) < b(n,j) \implies 1 < \frac{b(n,j)}{b(n,j-1)} \implies 1 < \frac{p}{1-p}\frac{\binom{n}{j}}{\binom{n}{j-1}}$$

$$b(n,j+1) < b(n,j) \implies \frac{b(n,j+1)}{b(n,j)} < 1 \implies \frac{p}{1-p} \frac{\binom{n}{j+1}}{\binom{n}{j}} < 1$$

But I'm still not sure where to go from here. Do you think you could give me another hint in the right direction?

 

[Ray Vickson]

Well, if I let $j$ be such that $b(n,j)$ is maximal, then

$$b(n,j-1) < b(n,j) \implies 1 < \frac{b(n,j)}{b(n,j-1)} \implies 1 < \frac{p}{1-p}\frac{\binom{n}{j}}{\binom{n}{j-1}}$$

$$b(n,j+1) < b(n,j) \implies \frac{b(n,j+1)}{b(n,j)} < 1 \implies \frac{p}{1-p} \frac{\binom{n}{j+1}}{\binom{n}{j}} < 1$$

But I'm still not sure where to go from here. Do you think you could give me another hint in the right direction?

Sure. Work at simplifying the ratio.

RGV

 

[jgens]

Okay, so I think I've got it. Rather than TeX the whole thing out right now, is the answer $j < p(n+1) < j+1$ correct?

 

[Ray Vickson]

Okay, so I think I've got it. Rather than TeX the whole thing out right now, is the answer $j < p(n+1) < j+1$ correct?

OK, if you replace $<$ by $\leq$.

RGV