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Let $S_n$ be defined as $\frac{1}{n}\sum_{t=1}^{t=n} [x^2+(p-q)x]$ where $x = 1-(1-p-q)^t$. We want to find the conditions on $p$ and $q$ such that $S_n$ is monotonically decreasing for all $n$. $0 < p,q < 1$ and $0 < 1-p-q < 1$.

Note: Till now I have tried to get a closed bound expression for $S_n$ and differentiate it w.r.t. $n$ to get the conditions for a negative slope but it is getting really complex.

Another approach was to reduce this expression to the sum $S_n = C (>0) + \frac{1}{n}\left[q\sum_{t=1}^{n} \lambda^{2t} + (p-q) \sum_{t=1}^{n} \lambda^t \right]$ where $\lambda=1-p-q$. We know the upper bounds of the two sums, and since the denominator grows more rapidly than the numerator, it is sufficient to show that the numerator is positive to get a monotonically decreasing sequence.

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  • $\begingroup$ $p\geq q$ is a sufficient condition. $\endgroup$
    – Ron P
    Sep 6, 2020 at 10:34
  • $\begingroup$ If $p\leq q$, you can rewrite the questions with $a=p+q$, and $b=q-p$, and the condition $1\geq a\geq b\geq 0$. $\endgroup$
    – Ron P
    Sep 6, 2020 at 10:41

1 Answer 1

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$S_n$ is always increasing in $n$. Note that $x$ is increasing in $t$; therefore, if $p\geq q$, then $x^2+(p-q)x$ is increasing in $t$.

If $q > p$, then you can set $a=1-p-q$, and $b=1-p+q$. The condition is $0<a\leq b<1$, and $x^2+(p-q)x=x(x-1+b)=(1-a^t)(b-a^t)$, which is increasing in $t$.

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  • $\begingroup$ Hi Ron, Thanks for the insights. I can't see how you have taken the $\frac{1}{n}$ factor into account. $x^2 + (p-q)x$ is increasing in $t$, therefore its summation is also increasing in $t$ but won't the $n$ factor in the denominator attenuate it? $\endgroup$ Sep 6, 2020 at 12:41
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    $\begingroup$ when you have a (finite) set of numbers and you add to it another number which is greater than all the others, then the average increases. $\endgroup$
    – Ron P
    Sep 6, 2020 at 12:52

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