See: http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)
The MGF is exp[lambda*(e^t - 1)].
It is exp(20t + 25/2*t^2).
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.
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)]
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.
A moment generating function does exist for the hypergeometric distribution.
See: http://en.wikipedia.org/wiki/Uniform_distribution_(continuous)
The MGF is exp[lambda*(e^t - 1)].
Using the Taylor series expansion of the exponential function. See related links
It is exp(20t + 25/2*t^2).
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.
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.
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)]
Your question did not identify one distribution in particular. I have provide in the related link the moment generating functions of various probability distributions.
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.
[(1 - p)/(1 - pet)]r for t < -ln(p) where p = probability of success in each trial, r = number of failures before success.