Moments with Laplace

The Endeavour 2024-11-04

This is a quick note to mention a connection between two recent posts, namely today’s post about moments and post from a few days ago about the Laplace transform.

Let f(t) be a function on [0, ∞) and F(s) be the Laplace transform of f(t).

$F(s) = \int_0^\infty e^{-st} f(t) \,dt$

Then the nth moment of f,

$m_n = \int_0^\infty t^n \, f(t)\, dt$

is equal to then nth derivative of F, evaluated at 0, with an alternating sign:

$(-1)^n F^{(n)}(0) = m_n$

To see this, differentiate with respect to s inside the integral defining the Laplace transform. Each time you differentiate you pick up a factor of −t, so differentiating n times you pick up a term (−1)ⁿ tⁿ, and evaluating at s = 0 makes the exponential term go away.

Brief outline of Laplace transforms (pdf)
Normal approximation to Laplace distribution?
Computing moments with a fold

The post Moments with Laplace first appeared on John D. Cook.

Moments with Laplace

The Endeavour 2024-11-04

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