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MX,Y(s,t) = exp{μxs + μYt + ½(σX2s2 + 2ρσXσYst + σY2t2)} Where X ~ N (μx , σX) and Y ~ N (μY , σY). Also Corr(X,Y) = Cov (X,Y)/{Var(X) . Var(Y)} = ρ

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Q: Moment generating function of a bivariate normal distribution?
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How do you derive the moment generating function of an exponential distribution?

To derive the moment generating function of an exponential distribution, you can use the definition of the moment generating function E(e^(tX)) where X is an exponential random variable with parameter λ. Substitute the probability density function of the exponential distribution into the moment generating function formula and simplify the expression to obtain the final moment generating function for the exponential distribution, which is M(t) = λ / (λ - t) for t < λ.


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A moment generating function does exist for the hypergeometric distribution.


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See: http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)


Obtaining moment generating function of poisson distribution?

The MGF is exp[lambda*(e^t - 1)].


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Using the Taylor series expansion of the exponential function. See related links


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It is exp(20t + 25/2*t^2).


Inverse Gaussian distribution-discuss moment and maximum likelyhood estimation?

I've included two links. The MLE of parameters of NIG distribution is the subject of current research as attached. The moment generating function is provided in the first link.


How do you derive Moment generating function of Pareto distribution?

The moment generating function for any real valued probability distribution is the expected value of e^tX provided that the expectation exists.For the Type I Pareto distribution with tail index a, this isa*[-x(m)t)^a*Gamma[-a, -x(m)t)] for t < 0, where x(m) is the scale parameter and represents the least possible positive value of X.


Moment generating and the cumulant generating function of poisson distribution?

The moment generating function is M(t) = Expected value of e^(xt) = SUM[e^(xt)f(x)] and for the Poisson distribution with mean a inf = SUM[e^(xt).a^x.e^(-a)/x!] x=0 inf = e^(-a).SUM[(ae^t)^x/x!] x=0 = e^(-a).e^(ae^t) = e^[a(e^t -1)]


Solutionsn of Moment-generating functions?

Your question did not identify one distribution in particular. I have provide in the related link the moment generating functions of various probability distributions.


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There are many, many formulae:for different probability distribution functions,for cumulative distribution functions,for moment generating functions,for means, variances, skewness, kurtosis and higher moments.There are many, many formulae:for different probability distribution functions,for cumulative distribution functions,for moment generating functions,for means, variances, skewness, kurtosis and higher moments.There are many, many formulae:for different probability distribution functions,for cumulative distribution functions,for moment generating functions,for means, variances, skewness, kurtosis and higher moments.There are many, many formulae:for different probability distribution functions,for cumulative distribution functions,for moment generating functions,for means, variances, skewness, kurtosis and higher moments.


How do you obtain moment generating function Log-normal distribution?

You cannot because it does not exist.Although all the moments of the lognormal distribution do exist, the distribution is not uniquely determined by its moments. One of the consequences of this is that the expected values E[e^tX] does not converge for any positive t.