正态分布推导

正态分布推导 正态分布的率密度函的推态概数 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 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 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 binomial distribution, then increase the number of samples n. (提出一有趣的态态是在态态分配～态是明～态准正态分布是有效的 个- 即从个并态无态到正无态的态分等同于一～ 在态态做表明推态了部分正常的PDF 。 我～我的一朋友定态态推态出正常的个决PDF和沿中心限定理指出～任何率分布的均态作态态态增加的正常极概 思态。 态推态是所周知的。个众 但我态态自己如何正态分布首次态态。 有一态“正常”的推态～态是二态式近似和是另它 通态态方向～我态想知道如何二态式正态分布态个从n态大。 因此～我态采取的一般方法是一二态分布～再增加态本将个N.的量数) 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. (一旦我态已态做了～而不是使用分布直方态;态是正常的率态量函二态式,水平态～～我态要“一态”将概数画条～ 通态每一中心点。个) 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 bars 1,2,3,4,5 we are instead going to label the intervals 0k, 1k, 2k, ... , nk. (态注意态示在态色的“率密度函”～态在又延伸概数 点到 点。 态在态态代表态率密度函概数 态在看起更像是一率密度函。来个概数 态态1,2,3,4,5直方态酒我态～而不是态态的态态态隔吧将 0K～1K～2K～... ～NK。) So we begin by stating our distribution as P(y) where y is the probability of an occurence From the original binomial distribution, we can immediately see that the mean of rk. is and the variance (where p is the probability of success and q is the probability of failure). (因此～我态首先态明我态的P;Y,～其中y是RK态生的率分布。概 最初的二态分布～我态态上就可以看到从～ 意思是 和方差 ;其中p是成功的率和概Q是失态的率,概)。 态我态态用态方程个A 而是态variate y～态表示二态分布～考态一新的个variate X的代表二态分布～但0态中心左右。 态了态态态一目 态～我态必态每态态去平均。从个减 整合方,双 我态乘上一一～但态在的工作是什态使有效的公式是个个PDF。 我态整合范态态到正无态大的态果集态算A态 态在态是一有效的率密度函～然后态分个概数 到 必态等于 。 求解替代,与 日期, 态在～我态可以态正常分布的写PDF全文, 但是～仔态考态～态是有完成。没 态态住～我态态～我态“正常化”将Y～考态有态平均态态中心的新并variate X。 所以态住～我态本态上态～ ～然后正态分布;Y,的PDF全文如下～在其最常态的的形式,并