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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.
a simple model that describe the distribution of velocity of molecules in a gas with probabality concept. according to this model the probability of finding a certain velocity in a gas is proportional to e-E/KT, where E=1/2 mv2 for a free atoms of a gas.
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)]
It is the distribution given by: Pr(X = x) = e-5-5*(5.5)x/x! for x = 0, 1, 2, ...
The kurtosis of a distribution is defined as the fourth central moment divided by the square of the second central moment. Unfortunately, this browser converts Greek characters to the Roman alphabet so I cannot use standard forms of equations but: Suppose that for a random variable X, E(X) = m (mu) and E[(X - E(X))2] = V = s2 (sigma-squared) then Kurtosis = E[(X - E(X))4]/s4. Excess Kurtosis is then Kurtosis - 3. If excess kurtosis < 0 the distribution is platykurtic. They have a peak that is lower than the Normal: the peak is flat and broad. The tails of the distribution are narrow. Uniform distributions are platykurtic. A mesokurtic distibution has excess kurtosis = 0. The Gaussian (Normal) distribution - whatever its parameters - is mesokurtic. The binomial with probability of success close to 1/2 is also considered to be mesokurtic. If excess kurtosis is > 0 the distribution is leptokurtic. Leptokurtic distributions have a high and narrow peak. A good example is the Student's t distribution.