Im trying to find the modal estimator for a bernoulli distribution. If anyone can help me with this (or any distribution for that matter), it would be great. thanx in advance ;)
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Wikipedia reveals all.\n\nThe third link lists online statistics textbooks.  The Statistics Homepage is especially good.
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1. Go to Google at http://www.google.com/\n\n2. Enter into the search field:\n\nPenalized Maximum Likelihood Estimation in Logistic Regression\n\nor even\n\nFirth algorithm Penalized Maximum Likelihood Estimation in Logistic Regression\n\n3. Wow!\n\nThere are resources such as this from the Oxford journal Biometrika:\n\nhttp://biomet.oxfordjournals.org/cgi/content/abstract/69/1/123\n\nPenalized maximum likelihood estimation in logistic regression and discrimination \nJ. A. ANDERSON and V. BLAIR \n\nDepartment of Statistics, University of Newcastle upon Tyne Manchester\nDepartment of Statistical Unit, Statistical Unit, Christie Hospital Manchester \n\nMaximum likelihood estimation of the parameters of the binary logistic regression model for pr(H|x) is discussed with separate discussion of sampling from (i) the conditional distribution of H given x, (ii) the joint distribution of H and x, and (iii) the conditional distribution of x given H. Difficulties associated with continuous x in the latter sampling scheme are discussed. To avoid these, penalized maximum likelihood estimates are introduced, which give estimates of the logistic parameters and a nonparametric spline estimate of the marginal distribution of x. Extensions to multinomial logistic regression are outlined. \n\nKey Words: Binary logistic regression • Discrimination • Maximum likelihood estimation • Multinomial logistic regression • Nonparametric density estimation • Penalized maximum likelihood estimation • Spline function\n\nThere should be enough there to give you a sense of the answer.
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I'm not sure what exactly you want explained about them. A complete explanation requires more than can be provided on Y!A.\n\nBasically, the Confidence Interval is the set of population parameters for which your sample statistic would be a reasonably likely outcome. \n\nIf you knew the population parameter (which you don't) the reasonably likely outcomes would be near the population parameter. How near depends on 1. the level of confidence desired, and 2. the standard error of the samping distribution of the sample statistic.\n\nYou don't know the population parameter, nor the sampling distribution of the sample statistic, but you estimate them from your sample. \n\nThe CI tells you which population parameters are plausible values for the sample statistic you observed.\n\nIf this isn't clear, I don't know how to make it more clear in this format. Please feel free to contact me for clarification.
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Sure.  Try these:\n\nhttp://www.people.fas.harvard.edu/~rtmoore/GMMC/heckman.pdf\nhttp://www.nottingham.ac.uk/economics/leverhulme/research_papers/05_44.pdf\n\nAnd if those don't help, try this:\n\nhttp://www.google.com/search?num=20&hl=en&lr=lang_en&sa=X&oi=spell&resnum=0&ct=result&cd=1&q=%22Heckman+Selection+Model%22&spell=1
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It is actually binomial with parameter p=0.5 and n=400. Since 400 is quite a large number, u can approximate this by the Normal distribution with the same mean and variance, which are, np=400 and np(1-p)=100.\n\nX ~ N(200,100)\n\nu wanna test that if 216 is within 95% interval.\n\nVariance=100\nSDev=10\n\n216-200=16,\n16/stdev=1.6\n\nso, 216 is 1.6 stdev from the mean 200. which is within +/-1.96(95% confidence interval. hence, the null hy is accepted, with 95% significance level.\n\n\n\nbtw, drbinsomething's solution is equivalent to mine. he directly test P as a random variable while I tested mean nP as the random variable. He didn't mention it, but he took P to be asymptotically Normal as I did. '1.96' only applies when the distribution is Gaussian/Normal.
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try\nhttp://www.peterwebb.co.uk/probability.htm
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Try Google or Yahoo.  You'll find worked examples there.  I think it needs diagrams to understand; they and the accompanying explanations are too complex to attempt to reproduce here.
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Get some help here:\nhttp://en.wikipedia.org/wiki/Brownian_movement
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No problem \n\ngo to google and search under "Z table" +statistics. \n\nYou'll find a lot.
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try here:http://rds.yahoo.com/_ylt=A0oGkioZvFhFdAwAHMVXNyoA;_ylu=X3oDMTE2bWhpaGtwBGNvbG8DdwRsA1dTMQRwb3MDMwRzZWMDc3IEdnRpZANERlg1XzMw/SIG=1245uujf7/EXP=1163529625/**http%3a//www.josleys.com/articles/printgallery.htm\nhttp://rds.yahoo.com/_ylt=A0oGkioZvFhFdAwANMVXNyoA;_ylu=X3oDMTE2ZTZjN2I5BGNvbG8DdwRsA1dTMQRwb3MDNwRzZWMDc3IEdnRpZANERlg1XzMw/SIG=123rq1n1p/EXP=1163529625/**http%3a//www.diku.dk/topps/bibliography/2002.html\nhttp://rds.yahoo.com/_ylt=A0oGkmncu1hF.EgAtYJXNyoA;_ylu=X3oDMTE3aHFmdGVlBGNvbG8DdwRsA1dTMQRwb3MDMTAEc2VjA3NyBHZ0aWQDREZYNV8zMA--/SIG=141kp4fee/EXP=1163529564/**http%3a//www.cs.mcgill.ca/%7ehv/classes/MS.01.Fall/MSconcepts.pdf%23search='Modelling%2520%2526%2520Simulation'
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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.
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go to this site, its very good,\n\nit gives clear explanations and statistics
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I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
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What do you mean by waiting time W_n? Do you mean the time between the n-th and (n-1)-st events (jumps) or the time elapsed from t=0 up the n-th event (jump)? I assume that you mean the first (interarrival times and not arrival times).\n\nThen all the W_n's are exponentially distributed, and they are independent. Independence implies that you can ignore the expression behind the conditioning bar, and\n\nP(W_2 <= s | W_1 <= z) = P(W_2 <= s) = 1-exp(-lambda*s), by definition of the exponential distribution.\n\nThe distribution of the arrival time T_n (second interpretation above) would be Gamma(lambda,n), and then the solution would be more complicated because lacking independence, we could not ignore the expression behind the conditioning bar.\n\nADDED TO ORIGINAL ANSWER: Well, it's more complicated then. First let W_1 = X and W_2 = X+Y, where X and Y are the first and second interarrival times, respectively, which are independent and exponentially distributed with rate lambda.\n\nLet p = P(X+Y <= s | X <= t) = P(X+Y <= s and X <= t)/P(X<=t) =\n= P(A)/P(X<=t). (By the definition of conditional probability.)\nThen P(A) = E(P(A|X)). (This is called the tower property of conditional expectations). To compute this, first compute \nh(x) = P(A|X=x), then let P(A|X) = h(X), and finally, P(A) = E(h(X)).\n\nThus, h(x) = P(A|X=x) = P(X+Y <= s and X<=t | X=x) = \n= P(x+Y <= s and x<=t | X=x) = P(Y<=s-x and x<=t | X=x) = \n= P(Y<=s-x and x<=t) (By the independence of X and Y.)\n= P(Y<=s-x) if x<=t and 0 if x>t =\n= 1-exp(-lambda*(s-x)) if ((s-x)>=0 and x<=t) and 0 otherwise =\n= 1-exp(-lambda*(s-x)) if (x<=s and x<=t) and 0 otherwise =\n= 1-exp(-lambda*(s-x)) if x<=min(s,t) and 0 otherwise.\n\nNow let min(s,t)=u, that is, u=s if s<=t and u=t if s>t.\n\nThen P(A|X) = h(X) = 1-exp(-lambda*(s-X)) if X<=u \nand 0 otherwise, and \nP(A) = E(P(A|X)) = E(h(X)) = integral(0..u) of \n(1-exp(-lambda*(s-y))) * lambda * exp(-lambda*y) dy = \n= 1-exp(-lambda*u) - u*lambda*exp(-lambda*s). \n\n(Because if g is a function then \nE(g(X))=int(0..infinity) of g(y)*f(y)dy, \nwhere f is the density of X, and in this case, g(X)=0 if X>=u, so it suffices to integrate from 0 to u. \nThe density of X is  f(y)=lambda*exp(-lambda*y), because it is exponentially distributed. I computed the integral in Maple but you can also do it by hand; it's not hard although you have to work a little bit to get it.)\n\nFinally, p = P(A)/P(X<=t) = \n= (1-exp(-lambda*u)-u * lambda* exp(-lambda*s)) / \n/(1-exp-lambda*t). \n\n(= 1 - lambda*t*exp(-lambda*s) / (1-exp(-lambda*t)) if s>t.\nYou can check e.g. that p goes to 1 if s goes to infinity and p goes to P(Y<=s) if t goes to 0, which might increase your degree of confidence in the result.)\n\nThis is pretty hard stuff, I hope you can work yourself through it! Ask me if something is still unclear.\n\nADDED TO THE ORIGINAL ANSWER: I checked your solution thoroughly and I like it very much. I think it's a nice trick to use event counts instead of waiting times.\n\nA third possible solution would be that given W_1<=t, the conditional distribution of W_1 is uniform over (0,t), and the conditional distribution of W_2 is that of the sum of a uniform random variable over (0,t) and an exponential random variable with rate lambda (and the RV's in the sum are  independent). Then the conditional cdf of W_2 could be computed by using the convolution formula, and finally, you could plug s into the conditional cdf.
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a) Can't sell more than 4, so 1*p(selling 1) + 2*p(selling 2) + 3p(selling 3) + 4p(selling4).\n\nb) Find the number n so that p(x>n) is less than 0.05 for a Poisson "vif" mean 3.
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http://en.wikipedia.org/wiki/Central_limit_theorem\n\nhttp://www.rand.org/statistics/applets/clt.html
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the formula for sd.dev of binomial distribution is "\nsqrt(n*p*q).\ncan't understand the table so use the fomula urself.hope this helps