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Prove that for positive integers $m, n$ and two positive numbers $p, q$ satisfying $p+q = 1$ we have

$$(1 − p^n)^m + (1 − q^m)^n \ge 1.$$

Using binomial theorem, we have $1-C^m_1*p^n+C^m_2*p^2n-\cdots $ for $(1 − p^n)^m$. And same logic for the second term of the equation but they don't cancel each other out.

Is there anything I am doing wrong? (this is a practice from a probability class)

james black
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1 Answers1

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You can rewrite $q$ as $1-p$.

$$(1-p^n)^m+(1-(1-p)^m)^n\ge1$$

Since $m,n\ge1$, you can prove your statement true for $n=1$ and $m=1$, because the lowest values of $(1-p^n)^m+(1-(1-p)^m)^n$ occur when $n=m=1$.

The values have a minimum at $1$. Although $p$ might be a really small value, the effects of $p$ are overruled by the effects of $q$, or vise-versa, since $p$ is as far away from $0$ as $q$ is from $1$.

$$1-p+1-1+p\ge1$$

$$1\ge1$$

Therefore, this is true.

Yash Jain
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  • so this has nothing to do with bonomial? and is there a need to prove a^n>a^m if n>m? – james black Feb 07 '18 at 03:43
  • Do you want yours to relate to binomial expansion? In that case, I will delete my answer if it is not useful. @jamesblack – Yash Jain Feb 07 '18 at 03:44
  • oh no it doesnt have to be yours will work perfectly i am just curious if this can be solved by binomial? – james black Feb 07 '18 at 03:47
  • Actually I don't get it. Yes, when $m=n=1$ we get the lowest values but we haven't proved even that!!! – Ken Draco Feb 07 '18 at 03:48
  • How is it shown that the lowest values occur at n=m=1? – DanielWainfleet Feb 07 '18 at 03:48
  • which is i a confused about because p,q<1 then p^n is actually smaller as n gets bigger – james black Feb 07 '18 at 03:50
  • so how do we prove that? but the logic of the first step is good and then the second step wouldnt work out...becuase (1-p^n)^m gets smaller as m gets bigger so its not the lowest value at all – james black Feb 07 '18 at 03:50
  • I'm afraid you'll have to go through the tedious binomial. Substitute $q$ with $1-p$ as Yash Jain suggested. Sorry, I'm a bit rusty with Latex to do this task. It looks like I need to recollect a few things in Latex after my long absence. Or hopefully someone else will do it. It should not be too huge, it seems to me. But I guess you'll have to split the problem into two parts. When m is larger than n and vice versa, e.g. you'll be able to write n=m+k... – Ken Draco Feb 07 '18 at 03:59
  • @YashJain the lowest values ... occur when n=m=1 This is not obvious offhand, and could use some more elaboration. the effects of p are overruled by the effects of q, or vise-versa It's not clear what this is supposed to even mean. – dxiv Feb 07 '18 at 04:07
  • @jamesblack....But I guess you'll have to split the problem into two parts. When m is larger than n and vice versa, e.g. you'll be able to write n=m+k.. In fact these will be two symmetrical cases. So, you don't really need to do a tedious thing twice. – Ken Draco Feb 07 '18 at 04:10
  • It was my mistake to post that comment, it made no sense. – Yash Jain Feb 07 '18 at 04:11
  • whats latex and can you elaborate the case a little more the two cases thanks – james black Feb 07 '18 at 04:51
  • If you want to write "x^2", write it between two dollar signs (dollar sign)/x^2/(dollar sign).

    This should output $x^2$. @jamesblack

     – Yash Jain Feb 07 '18 at 04:52