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I want to get the characteristic function of the bivariate normal distribution when both means =0 and variances =1 .

I can't do the integration. In attach file i don't know how he reached the last integration.

Masteras

06-26-2010, 07:50 AM

I suggest you start with the characteristic function of the univariate normal and then go the bivariate, trust me in this one. A ggod guide could be Casella dn Burger Statistical Inference. Both in the ub=nivariate and bivariate you must do a trick. It is called fulfilment of the square, that is you must create the square in the parenthesis of the exponential by adding and subracting terms. is that clear?

Sorry,but i didn't understand what you mean .

Could you explain this trick to me,please?

Does fulfilment of the square means complete the square?

I can get characteristic to univariate ,but the problem in bivariate is there is a term containing (xy) .I can't act with it .It causes a trouble in the integral .

Masteras

06-26-2010, 12:39 PM

Does fulfilment of the square means complete the square?

yes, so you know the trick, you have to do that for the bivariate also. It is hard, but it needs trying.

This is my try.What shall i do next?

Masteras

06-26-2010, 01:53 PM

Listen, this requires a bit thought and i am not going to solve it. But, first of all forget the imaginary part. the normal has finite moments, so use the moment generating function, it will be the same without the i. second, you are good there, i suggest you find the moment generating function of the bivariate normal on the web so that you will have a clue where you want to end. What you must do is fill the square of the bivariate normal in the exponent and every time you add and subract something take the other outside the integral. Now, i repeat myself, you must make integral equal to 1, thereofore what is outside the integral will be the final version of what you are looking for. find the answer on the web, it will help you understand what is missing from the square filling.

Ok,Masteras.Thanks for all,I will looking for it again.

Masteras

06-26-2010, 03:28 PM

I have been there many times and I know I will keep up being in that situation. Try and when you find it you will feel good and then you will move on for the next one.

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