Finding the mean value of a Pareto Distribution.

58 Views Asked by At

The question:

Quality control experts estimate that the time (in years) until a specific electronic part from an assembly line fails follows a Pareto density of the form $f(x) = 3/x^{4}$ for $1 < x \leq{\infty}$.

What is the approximate mean failure time?

My current attempt starts with the following:

$E[X] = \int_{1}^{\infty}xf(x)dx = \int_{1}^{\infty}x3x^{-4}dx = -x^{-3}$.

Is this antiderivative correct? And given the bounds of $1$ to $\infty$, how can I determine an approximate mean time from this?

1

There are 1 best solutions below

0
On

You can also use the fact that for a nonnegative random variable $X$ $$ \mathbb E[X] = \int_0^\infty (1-F(x))\ \mathsf dx, $$ where $F$ is the cumulative distribution function of $X$. So here $$ \mathbb E[X] = \int_1^\infty x^{-3}\ \mathsf dx = \frac12. $$