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Q: Standard deviation of the multinomial distribution of Bernoulli variables? ( Answered,   4 Comments )
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 Subject: Standard deviation of the multinomial distribution of Bernoulli variables? Category: Science > Math Asked by: mariane0-ga List Price: \$4.00 Posted: 06 Jun 2005 12:09 PDT Expires: 06 Jul 2005 12:09 PDT Question ID: 530009
 ```Binomial distribution: N = number of trials p = number of positive outcomes q = number of negative outcomes = 1 - p P(X) = probability of X in N positive answers P(X) = (N! / (X! (N - X)!)) * ((p/N)^X) * ((q/N)^(N - X)) variance = sqrt (N * (p/N) * (q/N)) Multinomial Distribution N = number of trials p1 = number of outcomes which are of class 1 p2 = number of outcomes which are of class 2 (...) pm = number of outcomes which are of class m P(X1, X2, ..., Xm) = probability of getting X1 outcomes of class 1 and X2 outcomes of class 2, ..., and Xm outcomes of class m. P(X1, X2, ..., Xm) = (N! / Product(p1! * p2! * ... * pm!)) * ((p1/N)^X1) * ((P2/N)^X2) * ... * ((Pm/N)^Xm) This is an m+1 dimentional bell shaped surface. (Though in practice only points corresponding to integer values of X1, ... Xm exist). Please, what is it's variance and standard deviation? Thank you.```
 Subject: Re: Standard deviation of the multinomial distribution of Bernoulli variables? Answered By: mathtalk-ga on 06 Jun 2005 19:01 PDT
 ```Hi, mariane0-ga: Because the multinomial distribution has, presumably in contrast to a binomial distribution, more than two possible outcomes, it becomes necessary to treat the "expected value" (mean) of the distribution as a vector quantity, rather than a single "scalar" value. Similarly the appropriate notion of "variance" is a square matrix, which is called the covariance matrix. The diagonal entries of this matrix can be interpreted in terms of the individual "binomial" outcomes (eg. red vs. not red), and the off-diagonal entries in terms of corresponding pairs of binomial probabilities. See here for a concise summary: [Multinomial distribution -- Wikipedia] http://en.wikipedia.org/wiki/Multinomial_distribution What this entails for the analog of "standard deviation" is a bit harder to decide without more information about the intended application. If the outcomes are not "categorical" but instead quantitative outcomes, then we can total all the N trials and come up with a single number, which amounts to a sum of binomially distributed random variables. I would gladly provide further details, if that is what you have in mind. regards, mathtalk-ga```
 ```Hi mathtalk. Thank you very much for your answer, I'm trying to pay you. My old credit card has been replaced by a new one, which has the same number but an expiry date of 04:08 instead of 04:05. I tried updating my details but it says "Unable to process your request, please try again later". I tried several times. If you have any kind of priviledge access to google database, maybe you could do it. I think I might have hit some kind of security preventing people from modifying their credit information while some payement is pending, in order to avoid paying. Unfortunately, it is also preventing me from entering the correct expiry date! I feel bad about this, as I really would hate cheating you. Please tell me what you want me to do. Mary```
 ```Hi, Mary: I appreciate your concern, but actually you don't need to do anything to pay me. The credit card on file with Google Answers will be charged at the end of the month or whenever the amounts reach some threshold. In any case the amount you've offered will most likely wind up being charged to your account regardless of the new expiration date on the card. Let me know if you need Clarification about the categorical vs. quantitative outcomes mentioned above. regards, mathtalk-ga```
 ```Oh good. I'll wait to see what happens at the end of the month then. If you don't mind, I'll tell you whether I need any clarification once I've read the wikipedia. Their server is down for maintenance until 21:00 UTC, it says. regards, Mary```
 ```Hi Mathtalk, I finally got round to looking up the formula, and yes, it's what I needed, thank you very much. I am now trying to pay you... I copy-pasted below my dialog with google answer support, I don't know whether you'll find it funny or infuriating, though... You're very welcome to close the question and attempt to collect your money. Please let me know if you succeed at Mariane at no-log dot org. Best regards, Mary Hello, Thank you for your reply. We will look into this matter further. Sincerely, The Google Answers Team Original Message Follows: ------------------------ Thank you for your answer, but this is unacceptable as I ow money to mathtalk-ga, who has been really helpful to me and is rightly expecting to be paid for his trouble. The only thing which is needed in order for him to get paid is updating my credit card expiry date. Please do it, manually if necessary. Mary Answers-Support@google.com wrote: >> Hello Mary, >> >> Thank you for your reply. Unfortunately, you may be unable to continue >> using the Google Answers service. >> >> Sincerely, >> >> The Google Answers Team >> >> Original Message Follows: >> ------------------------ >> Thank you for your reply. >> >> I only have one credit card. What should I do? >> >> Mary >> >> >> Answers-Support@google.com wrote: >> > >>>>Hello, >>>> >>>>Thank you for your email. We are sorry you are experiencing this >>>>difficulty. We recommend that you try a different credit card. >>>> >>>>Sincerely, >>>> >>>>The Google Answers Team >>>> >>>>Original Message Follows: >>>>------------------------ >>>>Hi, >>>> >>>>I'm user "mariane0-ga". >>>> >>>>My old credit card has been replaced by a new one, which has >>>>the same number but an expiry date of 04:08 instead of 04:05. >>>> >>>>I had already typed that, and I thought it had been taken >>>>into account and saved in the database. >>>> >>>>Unfortunately, the database has now reverted to an expiry date >>>>of april 2005, so of course it doesn't work. >>>> >>>>Now when I try to update my details, it says "Unable to process >>>>request, please try again later", but I've tried at least 10 >>>>times. >>>> >>>>This is all the more embarrassing as I ow money to a researcher >>>>who has been really helpful. >>>> >>>>Please help. >>>> >>>>Mary, "mariane0-ga" >>>>```