1

Let $\mathbf{a}\triangleq(a_k)_{k=1,2,\ldots,K}$, where $0<a_1<a_2<\cdots<a_K<1$, $K=1,2,\ldots$.

Let $$\bar{a}(p)\triangleq\sum_{k=1}^{K}\binom{K}{k}p^k(1-p)^{K-k}a_k$$ where $0\le p\le 1$.

Statement: $\bar{a}(p)$ increases with $p$.

If it is right, will you please help me to prove it? Many thanks for your help!

Zev Chonoles
  • 129,973
Dave
  • 576
  • This seems to belong in math.SE – Feyre Sep 13 '16 at 14:52
  • I do not think your statement is correct, for instance, $0<\bar{a}(1/2)>\bar{a}(1)\equiv0$ –  Sep 13 '16 at 14:59
  • When $p=1$, $\bar{a}(1)=p^Ka_K$. – Dave Sep 13 '16 at 15:07
  • Introducing $a_0=0$, one sees that $\bar a(p)=E(a_{S_p})$ where each random variable $S_p$ is binomial $(K,p)$. One can construct the family $(S_p)$ in such a way that $P(S_p<S_q)=1$ for every $p<q$, then the fact that the sequence $(a_k)$ is increasing shows that $P(a_{S_p}\leqslant a_{S_q})=1$ hence, indeed, $\bar a(p)\leqslant\bar a(q)$ (and the same idea, with more care, also shows that $\bar a(p)<\bar a(q)$). – Did Sep 13 '16 at 16:18
  • @Did, I very appreciate your kindly help. Thank you! – Dave Sep 14 '16 at 04:56
  • *Typo: ...in such a way that $P(S_p\leqslant S_q)=1$ for every $p<q$... – Did Sep 14 '16 at 05:32

0 Answers0