正態(tài)分布推導(dǎo)

1、正態(tài)分布的概率密度函數(shù)的推導(dǎo)An interesting question was posed in a Statistics assignment which was to show that the standard normal distribution was valid - ie the integral from negative infinity to infinity equated to one and in doing so showed the derivation of the part of the normal pdf  .A friend of

2、 mine and I decided to try to derive the normal pdf and the thinking went along the lines of the central limit theorem which states that the mean of any probability distribution becomes normal as the number of trials increases.The derivation of this is well known. but we asked ourselves how the

3、 normal distribution was first achieved. There is another 'normal' derivation which is the binomial approximation and it is through this direction that we wondered how to derive the normal distribution from the binomial as n gets large.So the general approach we will take is to take a b

4、inomial distribution, then increase the number of samples n.(提出一個有趣的問題是在統(tǒng)計分配，這是表明，標(biāo)準(zhǔn)正態(tài)分布是有效的 - 即從負無窮到正無窮的積分等同于一個，并在這樣做表明推導(dǎo)了部分正常的PDF  。我，我的一個朋友決定嘗試推導(dǎo)出正常的PDF和沿中心極限定理指出，任何概率分布的均值作為試驗增加的正常思維。這個推導(dǎo)是眾所周知的。 但我們問自己如何正態(tài)分布首次實現(xiàn)。 有另一種“正常”的推導(dǎo)，這是二項式近似和它是通過這個方向，我們想知道如何從二項式正態(tài)分布為n變大。因此，我們將采取的一般方法是

5、一個二項分布，再增加樣本N.的數(shù)量)Once we have done this, instead of using the horizontal lines of the distribution histogram (which would be the normal probability mass function of the binomial), we are going to 'draw' a line through each central point.(一旦我們已經(jīng)做了，而不是使用分布直方圖（這將是正常的概率質(zhì)量函數(shù)二項式）水平線，我們要“畫一條線”，通過每

6、一個中心點。)Notice how the 'probability mass function' shown in blue now extends from  point through to the  point. This probability mass function now represented by the blue line now looks more like a probability density function. Instead of labeling the histogram b

7、ars 1,2,3,4,5 we are instead going to label the intervals 0k, 1k, 2k, . , nk.(請注意顯示在藍色的“概率密度函數(shù)”，現(xiàn)在又延伸  點到  點。 現(xiàn)在藍線代表這概率密度函數(shù)現(xiàn)在看起來更像是一個概率密度函數(shù)。 標(biāo)簽1,2,3,4,5直方圖酒吧我們，而不是將標(biāo)簽的時間間隔0K，1K，2K，. ，NK。)So we begin by stating our distribution as P(y) where y is the probabilit

8、y of an occurence of rk. From the original binomial distribution, we can immediately see that the mean is  and the variance  (where p is the probability of success and q is the probability of failure).(因此，我們首先說明我們的P（Y），其中y是RK發(fā)生的概率分布。 從最初的二項分布，我們馬上就可以看到，意思是  和方差 （其中p是成功的概率和Q是失敗的概率）)。讓我們調(diào)用這個方程A而是讓variate y，這表示二項分布，考慮一個新的variate X的代表二項分布，但0為中心左右。 為了實現(xiàn)這一目標(biāo)，我們必須從每個值減去平均。 整合雙方： 我們乘上一個一個，但現(xiàn)在的工作是什么使有效的公式是PDF。 我們整合范圍負到正無窮大的結(jié)果集計算A值現(xiàn)在這是一個有效的概率密度函數(shù)，然后積分  到  必須等于  。求解與替代： 日期：現(xiàn)在