# 6.5. Odds Ratios¶

Binomial $$(n, p)$$ probabilities involve powers and factorials, both of which are difficult to compute when $$n$$ is large. This section is about simplifying the computation of the entire distribution. The result also helps us understand the shape of binomial histograms.

## 6.5.1. Consecutive Odds Ratios¶

Fix $$n$$ and $$p$$, and let $$P(k)$$ be the binomial $$(n, p)$$ probability of $$k$$. That is, let $$P(k)$$ be the chance of getting $$k$$ successes in $$n$$ independent trials with probability $$p$$ of success on each trial.

The idea is to start at the left end of the distribution, with the term

$P(0) ~ = ~ (1-p)^n$

Then we will build up the distribution recursively from left to right, one possible value at a time.

To do this, we have to know how the probabilities of consecutive values are related to each other. For $$k \ge 1$$, define the $$k$$th consecutive odds ratio

$R(k) = \frac{P(k)}{P(k-1)}$

These ratios help us calculate $$P(k)$$ recursively.

\begin{split} \begin{align*} P(0) &= (1-p)^n \\ \\ P(1) &= P(0) \cdot \frac{P(1)}{P(0)} = P(0)R(1) \\ \\ P(2) &= P(1)R(2) \end{align*} \end{split}

and so on.

Even though we already have a formula for the binomial probabilities, building the distribution using consecutive ratios is better computationally and also helps us understand the shape of the distribution.

## 6.5.2. Binomial Consecutive Odds Ratios¶

How is this more illuminating than plugging into the binomial formula? To see this, fix $$k \ge 1$$ and calculate the ratio $$R(k)$$.

\begin{split} \begin{align*} R(k) &= \frac{\binom{n}{k}p^k(1-p)^{n-k}} {\binom{n}{k-1}p^{k-1}(1-p)^{n-k+1}} \\ \\ &= \frac{n-k+1}{k} \cdot \frac{p}{1-p} ~~~ \text{(after cancellation)} \end{align*} \end{split}

Notice that the formulas for $$R(k)$$ are simple. This makes it easy to compute $$P(k)$$ recursively. For example, if $$n \ge 3$$, we can compute $$P(3)$$ as

$P(3) = (1-p)^n \big{(} \frac{n - 1 + 1}{1} \cdot \frac{p}{1-p} \big{)} \big{(} \frac{n - 2 + 1}{2} \cdot \frac{p}{1-p} \cdot \big{)} \big{(} \frac{n - 3 + 1}{3} \cdot \frac{p}{1-p} \cdot \big{)}$

Quick Check

In the binomial $$(60, 2/3)$$ distribution, $$P(35) = 0.04207955004383654$$. Use this (but no scipy or combinatorics) to find $$P(36)$$.

## 6.5.3. Shapes of Binomial Histograms¶

Now observe that comparing $$R(k)$$ to 1 tells us whether the histogram is going up, staying level, or going down at $$k$$.

\begin{split} \begin{align*} R(k) > 1 &\iff P(k) > P(k-1) \\ R(k) = 1 &\iff P(k) = P(k-1) \\ R(k) < 1 &\iff P(k) < P(k-1) \end{align*} \end{split}

Note also that the form

$R(k) = \big{(} \frac{n+1}{k} - 1 \big{)} \cdot \frac{p}{1-p}$

tells us the the ratios are a decreasing function of $$k$$. In the formula, $$n$$ and $$p$$ are the parameters of the distribution and hence constant. It is $$k$$ that varies, and $$k$$ appears in the denominator.

This implies that once $$R(k) < 1$$ for some $$k$$, it will remain less than 1 for all larger $$k$$. In other words, once the histogram starts going down, it will keep going down. It cannot come back up again.

That is why binomial histograms are either non-increasing or non-decreasing, or they go up and come down. But they can’t look like waves on the seashore. They can’t go up, come down, and go up again.

Let’s visualize this for a $$n = 23$$ and $$p = 0.7$$, two parameters that have no significance other than being our choice to use in this example.

n = 23
p = 0.7
k = range(n+1)
bin_23_7 = stats.binom.pmf(k, n, p)
bin_dist = Table().values(k).probabilities(bin_23_7)
Plot(bin_dist) # It is important to define k as an array here,
# so you can do array operations
# to find all the ratios at once.
k = np.arange(1, n+1, 1)
((n - k + 1)/k)*(p/(1-p))

array([53.66666667, 25.66666667, 16.33333333, 11.66666667,  8.86666667,
7.        ,  5.66666667,  4.66666667,  3.88888889,  3.26666667,
2.75757576,  2.33333333,  1.97435897,  1.66666667,  1.4       ,
1.16666667,  0.96078431,  0.77777778,  0.61403509,  0.46666667,
0.33333333,  0.21212121,  0.10144928])


What Python is helpfully telling us is that the invisible bar at 1 is 53.666… times larger than the even more invisible bar at 0. The ratios decrease after that but they are still bigger than 1 through $$k = 16$$. The histogram rises till it reaches its peak at $$k = 16$$. You can see that $$R(16) = 1.1666 > 1$$. Then the ratios drop below one, so the histogram starts going down.

## 6.5.4. Mode of the Binomial¶

A mode of a discrete distribution is a possible value that has the highest probability. There may be more than one such value, so there may be more than one mode.

We have seen that once the ratio $$R(k)$$ drops below 1, it stays below 1, so the histogram keeps falling. To identify the mode, therefore, we will find all values of $$k$$ such that $$R(k) \ge 1$$.

Let $$q = 1-p$$. Every value $$k$$ for which $$R(k) \ge 1$$ must satisfy

$\big{(} \frac{n+1}{k} - 1 \big{)} \frac{p}{q} ~ \ge ~ 1$

That is,

$\frac{n+1}{k} ~ \ge ~ \frac{q}{p} + 1 ~ = ~ \frac{1}{p}$

which is equivalent to

$k ~ \le ~ (n+1)p$

We have shown that for all $$k$$ in the range 0 through the integer part of $$(n+1)p$$, the histogram rises; for larger $$k$$, it falls.

Therefore the peak of the histogram is at the largest $$k$$ in this range. That’s the integer part of $$(n+1)p$$.

So the integer part of $$(n+1)p$$ is a mode of the binomial.

Because the odds ratios are non-decreasing in $$k$$, the only way in which there can be more than one mode is if there is a $$k$$ such that $$R(k) = 1$$. In that case, $$P(k) = P(k-1)$$ and therefore both $$k$$ and $$k-1$$ will be modes. To summarize:

The mode of the binomial $$(n, p)$$ distribution is the integer part of $$(n+1)p$$. If $$(n+1)p$$ is an integer, then $$(n+1)p - 1$$ is also a mode.

To see that this is consistent with what we observed in our numerical example above, let’s calculate $$(n+1)p$$ in that case.

(n+1) * p

16.799999999999997


The integer part of $$(n+1)p$$ is 16, which is the mode that we observed.

But in fact, $$np$$ is a more natural quantity to calculate. For example, if you are counting the number of heads in 100 tosses of a coin, then the distribution is binomial $$(100, 1/2)$$ and you naturally expect $$np = 50$$ heads. You don’t want to be worrying about $$101 \times (1/2)$$.

In fact you don’t have to worry when $$n$$ is large, because then $$np$$ and $$(n+1)p$$ are pretty close. In a later section we will examine a situation in which you can use $$np$$ to get an approximation to the shape of the binomial distribution when $$n$$ is large.