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Given $p \in [0,1], pn^{(d+1)/d} \to \infty$ and $np - \text{log}n - d \text{loglog}n \to \infty$, for a fixed d, show that $$n {{n-1} \choose d} p^d (1-p)^{n-1-d} \leq o(1), \text{as } n \to \infty.$$

I’ve been stuck with this for a day. Taking log of the first expression for $p$ would give log$n$, but I can’t see where that loglog$n$ comes from.

fresh_start
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2 Answers2

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From $\binom{n}{k}=\frac{n}{k}\cdot \binom{n-1}{k-1}$, we have $$\color{blue}{n\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}}=\\ (d+1)\cdot\frac{n}{d+1}\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}=\\ \color{red}{\frac{d+1}{p}}\cdot\color{orange}{\binom{n}{d+1}\cdot p^{d+1}(1-p)^{n-(d+1)}}\tag{1}$$ Of course $p\ne0$, because $pn^{(d+1)/d} \to \infty$ (otherwise it would $\to 0$).

Using binomial theorem $$\color{orange}{\binom{n}{d+1}\cdot p^{d+1}(1-p)^{n-(d+1)}}\leq\\ \sum\limits_{k=0}^n\binom{n}{k}\cdot p^k(1-p)^{n-k}=\\ (p+1-p)^n=1\tag{2}$$

Adding $(1)$ and $(2)$ altogether

$$\color{blue}{n {{n-1} \choose d} p^d (1-p)^{n-1-d}} \leq \color{red}{\frac{d+1}{p}}\tag{3}$$ The RHS is a constant, so $$n {{n-1} \choose d} p^d (1-p)^{n-1-d}=\mathcal{O}(1)$$

Update. However, the question is about $\mathcal{0}(1)$. Let's note by $b=\max\{p,1-p\}\in(0,1)$. From $(1)$ $$\color{blue}{n\cdot \binom{n-1}{d}\cdot p^d (1-p)^{n-1-d}}\leq\\ \frac{d+1}{p}\cdot\binom{n}{d+1}\cdot b^{d+1}b^{n-(d+1)}= \frac{d+1}{p}\cdot\binom{n}{d+1}\cdot b^{n}\leq \\ \frac{d+1}{p}\cdot n^{d+1}\cdot b^{n}= \frac{d+1}{p}\cdot \frac{n^{d+1}}{\left(\frac{1}{b}\right)^{n}}=...$$ noting $a=\frac{1}{b}>1$ $$...=\frac{d+1}{p}\cdot \frac{n^{d+1}}{a^{n}}\to0, n\to\infty$$ This post explains why.

rtybase
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  • Thanks for the answer. It seems like the given assumption was not used? – fresh_start Aug 02 '21 at 20:14
  • Not directly. Indirectly, to conclude that $p\ne 0$. In any case $pn^{\frac{d+1}{d}}>pn \to\infty$ is obvious for $p>0$. And $$np - \log{n} - d \log\log{n}= np-\log{\left(n\cdot\log^d{n}\right)}=\ \log{\left(\frac{e^{pn}}{n\cdot\log^d{n}}\right)}> \log{\frac{e^{pn}}{n^{d+1}}}\to\infty$$ too. – rtybase Aug 02 '21 at 20:22
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Since $d$ is fixed, $0<p<1$ we can estimate $$ n {{n-1} \choose d} p^d (1-p)^{n-1-d} \leq c\,n^{d+1}(1-p)^{n}, $$ which tends to zero, because $(1-p)^n$ exponentially small. If I understand your problem correctly.

vesszabo
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  • But how do you show $(1-p)^n$ tends to $0$ fast enough to cancel out the $n^{d+1}$? – fresh_start Jul 23 '21 at 11:22
  • @JoshNg Choose a positive integer $k$ such that $d+1\leq k$, and $1-p=\frac{1}{b}$, where $b>1$. So it is enough to prove that $\frac{n^k}{b^n}$ tends to zero. – vesszabo Jul 23 '21 at 11:28