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1、正态分布的概率密度函数的推导An interesting question was posed in a Statistics assignment which was to show that the standard normal distribution was val id ie the integral from negative infinity to infinity equated to one and in doing so showed the deri vat ion of the part of the normal pdf.A friend of mine and I
2、 decided to try to derive the normal pdf and the thinking went along the lines of the central 1 imit theorem which states that the mean of any probabi 1 ity distribution becomes normal as the number of trials increases.The derivation of this is wel 1 known. but we asked ourselves how the notnvil dis
3、tribution 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
4、 increase the number of samples n.（提出一个有建的问题是在统计分配.这是表明，标准正态分布是有效的-即从负无穷到正无穷的积分等同于一个，井在这样做表明推导了部分正常的PDF 啊亍o我,我的一个用友决定券试推导出正常的PDF和沿中心极限定理指出，任何概率分布的均值作为试验增加 的正常思维.这个推导是众所周知的。但我们问自己如何正态分布首次实现。有另一种“正常”的推导.这是二项 式近似和它是通过这个方向，我们想知道如何从二项式正态分布为n变大 因此，我们将采取的一般方法是一个二项分布，再增加样本&的数量）X)Once we have done this. inst
5、ead of using the horizontal 1 ines of the distribution histogram (which would be the normal probability mass function of the binomial) f we are going to * draw * a line through each central point.（一旦我们巳经做了,而不是使用分布直方图（这将是正常的歧率质*函教二项式）水平线，我们要“画 一条线”，通过每一个中心点。）ot ice how the 1 probabi 1 i ty mass funct
6、ion shown in blue now extends from A = () point through to the A = 5 point. This probabi 1 ity mass function now represented by the blue 1 inc now looks more 1 ike a probabi 1 ity density function. Instead of labeling the histogram bars 1 . 2.3. 1.5 we are instead going to label the intervals Ok. Ik
7、. 2k nk.G#注示在黄色的密度居數.现在又楚伸 X =0点到 X = 5也 ，：歸.， 这從率密度函数现在看起来更像是一个概率密度函数C标备1.2.3.4.54方图酒吧我们，而不是将标签的时间间隔OK. IK, 2K NK.）X)So we begin by stating our distribution as P(y) where y is the probab i 1 i ty of an occurcnce ofrk. From the original binomial distribution, we can immediately see that the meanis=U
8、pk and the variance(where p is the probabi 1 i ty of success and q is the probabi1ity of failure).（因此.我们首先说明我们的P （Y）.其中y是RK发生的慨宰分布. 从最初的二项分布，我们马上就可 以看到，意思是M = 和方差a2 = npqk（其中p是成功的宰和q是失败的帙率）.?y = P(Y=rk)=(k=(宀)bsincr wo arc going to be dcmring the formula of the gradient function we arc going to have
9、 to derive the formula for the gradient from first principles.#=P(Y=(r+l)k)_ ( t4- 1 n r1 (As a liint to wliero we are going, we eventually want to get to (1/ n!n!r 7tr、r!(n r)!)n!(r + 1) 宀ir+l nr-l(r + l)!(n - r - 1)! 咆卫.)广矿宀 . (_ 皿 (r + l)!(n - r)! J(r + l)!(n r)!)(邛(二1广T+MLn!ow w? arc ready to di
10、vide by yU _y _ 二；【PS一T)p一(r + 1 用 y (朮与)ppiWhiuh can be simplified aft er lots of cancellation to:=r(n .) . (r + 1”一 (r+ l)qAnd since p+ =1=詳伊-*)让我们调用这个方程A而是让variate y.这表示二项分布考虑一卜新的variate X的代表二项分布，但0为中心左右。为了实现这一目标，我们必须从每个值编去平均=rk=k(r np)r = - + np krt = x 4- upAnd:r + 1 = 7 + np + 1 kk(r + 1)=工 +
11、k np + kMultiply sides by kqkq(r + 1) = kqx + k2 npcf + A，q=(a: + knp 4 k)k(jNow taking Y|iiation A and multiplying top and bottom hy k2寸_ V 尸(叩_厂_ q) A,2(r + l)q k(k(iip r) kq) (r + knp 士 k)kq kq)x + knp 4- k)kq k(_T kq)k 2njxi + (x + k)kqSince k is the chango in x. k = dx and(y2 = k2njj(j. then:dy
12、 (- q dr )dx y a1 + (j: +(Lr)q diNow & we increase 11 to infinity anti dx tends to 0:机=*=A r.rp我们乘上一个一个,但现在的工作是什么使有效的公式是PDF。 我们整合范国负到正无穷大的结果集计 算A值 现在这是一个有效的做率密度函然后积分-oo到oo必须等于1 0心=130A capoo22匚（-名）&=!J 8N（w mnsifkr the following t nuiftfonxiation;experpctr d tjNow consilor:r2 =（+护）Tliercfcjre:But to convert into polar co-orclinatrst wo have to now apply the trnnsfoniiatioii求解与替代，日期:V现在，我们可以写为正常分布的PDF全文，焉（罚普）但是，仔细考虑.这是没有完成。请记住，我们说，我们将“正常化” Y.并考虑有关平均值为中心的新 rariate Xo 所以记住.我们本质 克, X = Y _卩， .分布（Y）的PDF金文如下，井在 其最常见的的形式：（素材和资料部分来自网络，供参考。可复制、编制，期待您的好评与关注）
