# Probability for a bin in a binned histogram

This question is very basic, but I cannot figure the error in my thinking. According to the author of the book "Pattern Recognition and Machine Learning", we can get the Probability Distribution Function of a distribution in the form of histograms by

"simply divide $$n$$ (the number of observation for one bin) by the total number $$N$$ of observations and by the width $$\Delta_i$$ of the bins to obtain probability values given by"

$$p(x) = \frac{n_i}{N\Delta_i}$$

I simply cannot get my head around how this give me the probability for a specific bin. $$N\Delta_i$$ is basically the sum of the area of every histogram. To calculate a relative frequency of one specific bin, why do we ignore the width $$\Delta_i$$ in the nominator, which would equal to the division of the area of one bin over the whole area.

• By the very definition of a PDF, the $p$ are not "probability values:" they estimate probability density, which is probability per unit width.
– whuber
Commented May 18, 2022 at 16:23
• I see, so basically as soon as I multiply $n_i$ by some width, I get the actual probability, which is the case I described. Commented May 18, 2022 at 16:30
• @kklaw Yes. You can further understand this in terms of integration, if you are familiar with calculus. Commented May 18, 2022 at 16:31
• – Tim
Commented May 18, 2022 at 16:50

## 1 Answer

One style of histogram of a sample has a vertical axis called Density, scaled so that the total area of the histogram bars is unity $$(1).$$ Thus, suppose you have a large sample from a population with density function $$f_X(x).$$ Then the histogram will tend to imitate the shape of $$f_X(x).$$ That is, the area of a histogram bar with base $$(a,b]$$ of width $$\Delta = b-a$$ will approximate $$P(a < X \le b) = \int_a^b f_X(x)\, dx.$$

For example, suppose x is a sample of size $$n = 1000$$ from a population distributed $$\mathsf{Gamma}(\mathrm{shape}=3, \mathrm{rate}=1/5).$$ Then we might have one of the two histograms shown below, each along with the density function $$f_X(x)$$ of $$\mathsf{Gamma}(3,1/5).$$ In this example, the population mean is $$\mu = 15, \sigma = \sqrt{75} \approx 8.660.$$ (Using R, where parameter prob=T of function hist plots a density histogram, and parameter br suggests the number of bins.)

set.seed(2022)
x = rgamma(1000, 3, .2)
summary(x)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
1.236   8.673  13.495  15.013  19.939  53.914
sd(x)
[1] 8.488901

par(mfrow=c(1,2))
hist(x, prob=T, br=5, ylim=c(0,.06), col="skyblue2")
curve(dgamma(x, 3, .2), add=T, lwd=2, col="brown")
hist(x, prob=T, ylim=c(0,.06), col="skyblue2")
curve(dgamma(x, 3, .2), add=T, lwd=2, col="brown")
par(mfrow=c(1,1))


In the left panel, the bin with base $$(10,20]$$ of width $$\Delta = 10$$ contains $$432$$ observations, has height $$.0432,$$ and thus area $$0.432.$$ According to the density function, the probability within this interval is $$0.4386.$$

diff(pgamma(c(10,20), 3,.2))
[1] 0.4385731


In the right panel, the bin with base $$(5,10]$$ of width $$\Delta = 5$$ contains $$432$$ observations, has height $$.0496,$$ and thus area $$0.248.$$ According to the density function, the probability within this interval is $$0.243.$$

diff(pgamma(c(5,10), 3,.2))
[1] 0.2430222


Note: In R, some details of a particular histogram can be listed by making a non-plotted histogram (parameter plot=F.) For the first histogram above, we have the following partial printout:

hist(x, prob=T, br=5, ylim=c(0,.06), plot=F)

$breaks [1] 0 10 20 30 40 50 60$counts
[1] 321 432 196  37  12   2

$density [1] 0.0321 0.0432 0.0196 0.0037 0.0012 0.0002$mids
[1]  5 15 25 35 45 55

...