# Please help me clarify one step in my statistics text that includes logarithm.

The following is from p. 103 of "An introduction to probability" Third Edition Volume I by William Feller:

The passage I was reading was about the derivation of Poisson Distribution, but I am stuck at this one step:

It starts with:
$$(n-v)^v < (n)_v < n^v$$

where $(n)_v$ is defined as:
$(n)_v=n(n-1)(n-2)\cdots(n-v+1)$ with $v\leqslant n$, both are positive integers.

Then, it proceeds to the following, which I am comfortable with:
$$n^v\left(1-\frac{v}{n}\right)^{v+r} < v!S_v < n^v\left(1-\frac{v}{n}\right)^r$$

where $S_v$ was given as:
$S_v={n \choose v}(1-\frac{v}{n})^r$ , the r is also positive integer.

Since, For $0<t<1$, $t<-\log(1-t)<t/(1-t)$, which is evident from the Logarithmic expansion... (The following line starts to throw me off)

$$\left( ne^{-(v+r)/(n-v)} \right)^v<v!S_v< \left( ne^{-r/n}\right)^v$$

I got lost when trying to get the result above, here are what I have been doing:

I tried to convert:
$t<-\log(1-t)<t/(1-t)$
Into:
$e^t(1-t)< 1 < e^{t/(1-t)}(1-t)$
Since I figured the middle '$v!S_v$' did not change, I tried to get the middle part to one.

However, I still can't match up $t$ with rest of the variables.

Any help or hint is appreciated.

## 2 Answers

The inequality $t < -\log\left(1-t \right)$ implies $r \log \left(1-t\right) < -r t$ and you can use this on the upper inequality (with $t=v/n$). A similar process works for the lower inequality.

For the answer I will be using 'ln' instead of 'log' for natural logarithm, since that is how I was taught. The following might be a bit too wordy for trained mathematicians. I write these down mainly as my personal records of working.

To start: $n^v (1-\frac{v}{n})^{v+r} = n^v e^{(v+r)ln(1-\frac{v}{n})}$

From: $t/(1-t) > -ln(1-t)$, sub t with v/n, We get:
$-\frac{v}{n-v}<ln(1-\frac{v}{n})$

Compare with the exponent above, it is useful to say:
$-(v+r)\frac{v}{n-v} < (v+r)ln(1-\frac{v}{n})$

Since $e^a<e^b$ when $a<b$:
$n^v e^{-(v+r)\frac{v}{n-v}}< n^v e^{(v+r)ln(1-\frac{v}{n})}$
Recall: $n^v (1-\frac{v}{n})^{v+r} = n^v e^{(v+r)ln(1-\frac{v}{n})}$

Thus: $n^v e^{-(v+r)\frac{v}{n-v}} < n^v (1-\frac{v}{n})^{v+r}$

Similarly, using:
$t<-ln(1-t)$, together with:
$n^v(1-\frac{v}{n})^r=n^ve^{r \cdot ln(1-\frac{v}{n})}$, we can get:
$n^v e^{r (-\frac{v}{n})} > n^v(1-\frac{v}{n})^r$

Finally, all I have to do is to link everything together:
$$n^v e^{-(v+r)\frac{v}{n-v}} < n^v (1- \frac{v}{n})^{v+r} < v! \cdot S_v < n^v(1-\frac{v}{n})^r < n^v e^{r (-\frac{v}{n})}$$

Finished.