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The derivative of the moment generating function and the mean

In this section we are making a connection by computing the mean of the Gumbel distribution based on the moment generating function (MGF). If you know the MGF, the integral below (which is also considered as the bilateral Laplace transform with a negative argument, $\mathcal{B}(-t)$)

In this section we are making a connection by computing the mean of the Gumbel distribution based on the moment generating function (MGF).

Background: the derivative of the moment generating function and it's relationship to the mean

If you know the MGF, the integral below (which is also considered as the bilateral Laplace transform with a negative argument, $\mathcal{B}(-t)$)

$$\begin{array}{rcl} M^\prime(t) &=& \int_{-\infty}^{\infty} x \underbrace{e^{tx}}_{\text{$e^{tx}= 0$ if $t=0$}} f(x) \,dx \\ \\ M^\prime(0) &=& \int_{-\infty}^{\infty} x f(x) \,dx = \mu \end{array}$$

$$\begin{array}{rcl} M^\prime(t) &=& \int_{-\infty}^{\infty} x \underbrace{e^{tx}}_{\text{$e^{tx}= 1$ if $t=0$}} f(x) \,dx \\ \\ M^\prime(0) &=& \int_{-\infty}^{\infty} x f(x) \,dx = \mu \end{array}$$

$$M(t) = \mathcal{B}(-t) = \int_{-\infty}^{\infty} e^{-tx} f(x) \,dx$$

$$M(t) = \mathcal{B}(-t) = \int_{-\infty}^{\infty} e^{tx} f(x) \,dx$$