Statistics: What is the Moment Generating Function of The Gamma Distribution?
824761
MGF of gamma is:\n\n1/(1-beta*t)^alpha\n\ncheck for yourself that the mean is:\n\nalpha*beta, and the variance is\n\nalpha*beta^2\n\nYou can derive the results yourself by the integration, but make certain you know how to integrate the gamma function (which is different from the gamma distribution).
135363
First, I agree with the answer that factorials are only define  for natural numbers.\n\nThere is however, the Gamma function from statistics.\n\nGamma(t) = (integral from 0 to infinity)(e^(-x)x^(t-1)dx\n\nHas the property that Gamma(n+1) = nGamma(n) = n!.
469250
Discrete Case:\nE[X] = Sigma_{i=1}^n x_i*p(X = x_i) \n\nwhere Sigma_{i=1}^n p(X = x_i) = 1\n\nContinuous Case:\nE[X] = int(x*p(X = x)dx,-infty,infty)\n\nThink of it as a weighted *average*.
177395
The easiest way to generate random number from a given distribution is to generate a uniform (0,1) random number; then compute the inverse cummulative probability function from that number:\nLet u be a uniform random number\nthen, x=Pinv(u,l) distributes Poisson if Pinv stands for the inverse function of the cummulative poisson distribution.\nIt's not relevant what language you use for programming
908707
a.\nThere are few observation in X(t)\n-constant w set\n-cos and sin are periodic functions, this means cos(t) and cos(t+T) would behave identical over the time period T, 2T,3T... [consider sameargument for sin too]\n\nFor your refference let us consider the indices as n=1,2,..\n\nSo, for each n, X(t) will have a joint distribution with independent unit normal random variables Y1,Y2. And, this would be nothing but linear combination of two unit notmal distirbutions.\n\nSince, the behaviour if sine and cosine remains same over time intervals (nT), this joint distribution will remain same for all n=1,2,...\n\nThis indicated that the process {X(t)} is stationary.\n\nSince, the mean & variance of the each distribution occuring at eact time interval may not remain constant or same, This is the reason why {X(t)} is a wekly stationary process.\n\nI hope now you understand.\n\nAll the best.
402618
I have no idea about the math, but I'm thinking the assumption is that the data are relatively correlated linearly.  There's a lot of math behind the wiki link.
710920
When working with data, skewness is the average of ((x_k-mu)/sigma)^3, where x_k are the elements of the data set, mu is the sample mean and sigma is the sample standard deviation.\n\nWhen working with random variables in probability, skewness is the mean (expected value) of ((X-mu)/sigma)^3, where X is the random variable, mu is its mean (expected value) and sigma is its standard deviation. The two definitions are analogous.\n\nThe normal distribution has skewness 0. In fact, all symmetric distributions have skewness 0. A positive skewness means that the distribution has a heavy tail to the right from the mean and it bends to the left and a negative skewness means the opposite.\n\nI'll give you a Wikipedia article although it wasn't written very well in my opinion, and I don't know if you can figure out anything from it.
755513
The t-distribution is related to the F-distribution as follows: the square of a value of t with n degrees of freedom is distributed as F with 1 and n degrees of freedom.\n\nSee for instance here:\nhttp://en.wikipedia.org/wiki/Student's_t-distribution
735064
That sounds familiar... Think I'm working on it now! But I don't remember q. It is n in my book. In which case, q (if I'm thinking right) is the number of samples. Then again what's up with the ^'s? If np was > 10, then it meant that the distribution was approx. normal. You might be ahead of me right now...
332575
A frequency distribution describes how often the potential responses occured.  The distribution might NOT be simple bell curve.\n\nIf you had a room full of people, and you counted how many of them weighed less than a 100 lbs, then you counted how many weighed between 100-110, then how many between 110-120, and kept counting until you had counted all the 10-lb groups, you would have a frequency distribution.\n\nThe distribution is normally displayed by making a bar graph with a bar for each group, in this example, a bar for each 10-lb group.  The bar is as tall as the number of people counted in that group.
1166976
As to part one, I can not recommend any acceptable, known test for such data (perhaps what I've written below might work).\n\nAs to part two, this is a horrid approximation since the midrange as the mean is a major assumption (even more far-reaching than assuming something about the populations' probability distributions).\n\nFor part 3 and also part 1:\n\nThe first problem you run into is that you do not know anything about the nature of the population distribution (normal, uniform, etc).  Thus, a nonparametric test would be required.  I know of no nonparametric tests for any of the statistics you have (minimum, maximum or range).\n\nIf it is possible to obtain the actual values for each of the samples in each set, the Kruskal-Wallis test should suffice.  It is nonparametric, so there are no problems with the requirement of assumptions about population distributions.\n\nFurther, its job is to determine if two samples are far enough apart such that the null hypothesis that the population distributions are the same can be rejected.  This would give you what you desire.\n\nAs far as your problem is concerned, the only statistics you have are minimum, maximum and range.   Perhaps one could derive a test to test the minima or maxima of each set, but one still runs into the problem that there can be no parameter placed on the population distribution.\n\nIf you can make an assumption about the distributions of the populations of interest (normal or whatnot), then it would be possible to derive such a test.  This would be a tedious task, however, as it would require thorough knowledge of order statistics and distribution functions.\n\nOne would need to use a test statistic based of off the difference in the two statistics that you use (I would use the maxima since they are further apart than the minima and probability distributions for ranges are uncharted territory unless you're a Ph.D.).  Thus, your test statistic is based off of the difference between the maxima (Y(n) - Z(n) where the (n) are subscripts, which is 2 in this case).\n\nTo perform this test, you would need to derive the probability distribution of the difference of the two maxima and use that to find the probability of the difference being two or greater.  This is quite a task for anyone and requires a large amount of knowledge in mathematical statistics.  I believe I could perform such a task, but it would likely take some time to complete.\n\nRemember though that you must make a critical assumption about the population before you can even do this.  I would recommend using the assumption of uniformity if you are unsure of the populations' distributions.  This is because the derived test would then be least likely to give a type I error, where you reject the data improperly, because the test is less likely to reject.
964589
to find the mean and variance of a MGF you need to find the moments.  To find the first moment, take the first derivative and then set t=0.  This would be (3 + 16t)*e^(3t + 8t^2).  With t=0 this equals 3.  The second moment is equated from the second derivative.  This is 16*e^(3t + 8t^2) + (e^(3t + 8t^2))*(3+16t)^2.  With t=0 the second moment equals 16 + 9 = 25.  \n\nThe mean is the first moment when t=0.  This equals 3.  \n\nThe variance is the second moment minus the first moment squared.  This equals 25 - 9 = 16.
165255
Take a look at the error function, and modify it so that you are dealing with distributions in 2 dimensions, using polar coordinates.\n\nFor example, the counterpart of the error function in 2 dimensions, in polar coordinates can actually be integrated, to produce the comulative distribution function\n\nF(R,s)=1-e^[-(R^2)/(2*s^2)]\n\nwhere s is the standard deviation, and R is the radius of a circle centered on the origin. \n\nAs R--> infinitity, then F(R,s)----> 1\nF(0,s)=0\n\nSuppose that the target has an edge R0 such that\n\nF(R0,s)=.5 for a given value of s. \n\nThen, you have to find the value of S for which F(R0,S)=0.9.
699053
The probability density function (pdf) of  Rayleigh Distribution is given as\n\n  p(r) = (r / q^2)exp [(-1/2) ( r^2 / q^2)]\n\nwhere q = standard deviation (r.m.s value) and  q^2 = variance or time averaged power of the signal.\n\nThe Cumulative Distribution Function (CDF) is given as \n\n P(R) = P(r <= R) = 1 - exp[(-1/2)(R^2 / q^2)]\n\nI have assumed that you are familiar with the concepts of Random Variables.
17841
You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
468863
Here's an article about it:\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
199925
Probability function of a poission distribution\n==> P(X=x) = e^(-m)*m^(x)/x! \n[x= 0,1,2,.... ]  countably infinite\n\nMean = 4 ==> m = 4\n\nProbability that x = 2\n==> P(X=2)= e^(-4)*4^(2)/2!\n          = \n\nProbability that x <= 2\n==> P(X<=2)= P(X=0) + P(X=1) + P(X=2)\n==> P(X<=2)\n= e^(-4)*4^(0)/0! + e^(-4)*4^(1)/1! + e^(-4)*4^(2)/2!\n= e^(-4) {1/1 + 4/1 + 16/2}\n= e^(-4) {1 + 4 + 8}\n= e^(-4) {13}\n\nProbability that x > 2\n==> P(X > 2)\n= P(X=3) + P(X=4) + ...... \n(Or) = 1 - {P(X<=2)}\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/