The classical statistical problem started with gambling, so let's look at the problem of gambling.
Q:A, b Two people for n innings gambling, a win probability is P, now set random variable x to show a win the number of innings, when X>np,a to the casino x-np yuan, otherwise b to the casino np-x yuan, then solve the casino earning expectations?
There are obviously two distributions (Bernoulli distributions) in this problem, but the dilemma we face is that this is a probability that a random variable x falls in a range based on a two-item distribution, and if we use the two-item distribution to multiply each other, we get an unusually tedious formula, which is extremely detrimental to the calculation.
To solve this problem, the mathematician thought of a method: it is well known that in continuous random variables we use the probability density curve f (x) and the area of the x-axis to characterize the probability of a set occurrence, here, the two-item distribution is a typical discrete distribution column, we make the column chart, the area of the column chart represents the probability. Then our two-item distribution of probability density "column chart", in fact, is to find an interval section of the column chart area and, at this time, the use of calculus tools. We can differentiate the curve into countless small rectangles, and of course, a series of small rectangles can fit a probability density curve. And this process uses the strling formula.
The following procedures are available:
Let's start with a special case.
It is easy to see that the probability density curve f (x) has been fitted out.
Therefore, we can easily solve the probability of distributing a section of two items by using the integral tool.
The above examined the situation of the P=1/2 in two distributions, the remaining problem is to promote, according to similar ideas and processes, we will get the di Moivre-Laplace limit theorem, which is the normal distribution of the original prototype:
Reference Blog: http://www.52nlp.cn/%E6%AD%A3%E6%80%81%E5%88%86%E5%B8%83%E7%9A%84%E5%89%8D%E4%B8%96%E4%BB%8A%E7%94%9F%E4%B8%80
"A first Course in probability"-chaper5-continuous random variable-normal distribution