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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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.
In statistics, the ogive curve is an approximation to the cumulative distribution function. It can be used to obtain various percentiles quickly as well as to derive the probability density function.
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The mode of the Pareto distribution is its lowest value.
To derive the mean of generalized Pareto distribution you must be good with numbers. You must be good in Calculus, Algebra and Statistics.
derive cost function from production function mathematically, usually done by utilizing mathematical optimization methods.
The total deviation from the mean for ANY distribution is always zero.
Here is the derivation on dsplog: http://www.dsplog.com/2008/07/17/derive-pdf-rayleigh-random-variable/
Here is qn excellent article that explains step by step: http://MasteringElectronicsDesign.com/how-to-derive-the-instrumentation-amplifier-transfer-function/
The probability mass function (pmf, you should know this) of the Poisson distribution isp(x)=((e-λ)*λx)/(x!), where x= 0, 1, ........Then you take the expected value of exp(tx), you should always keep in mind to find the moment generating function (mgf) you must always do(etx)*p(x), where t is a random variableTherefore,(etx)*((e-λ*λx)/(x!))(e-λ)*sum[(e-λ*λx)/(x!)]Thee-λ is only a constant; thus, it can be pulled out of the sums.Continuing,(e-λ)*sum[(λ*et)x)/x!]Let y=λ*et(e-λ)*sum[(y)x/x!]By Macalurins series, the sum[(yx)/x! ]= eySoonwards(ey)*(e-λ)Lets return the y by λ*et
var(X) = (xm/a - 1)2 a/a-2 . If a < or equal to 2, the variance does not exist.
The Gaussian distribution is the same as the normal distribution. Sometimes, "Gaussian" is used as in "Gaussian noise" and "Gaussian process." See related links, Interesting that Gauss did not first derive this distribution. That honor goes to de Moivre in 1773.