# Can the variance of a continuous random variable with known distribution be impossible to find? [duplicate]

This question already has an answer here:

I am solving a problem where the life expectancy of a microorganism can be modeled as having the PDF:

$f(x)= \left\{ \begin{array}{ll} kx^{-3} & x\geq 1 \\ 0 & x \lt 1 \\ \end{array} \right.$

Where $k=2$. I've calculated the average to be 2 hours, but when I try to calculate the variance as: $$\sigma^2=\int_1^\infty x^2f(x)dx - \mu^2$$

The integral doesn't converge. Am I doing something wrong? Thanks in advance!

## marked as duplicate by Glen_b♦ self-study StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 5 '17 at 0:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• It's absolutely possible that an expectation exists by a variance does not. – Matthew Drury Jun 4 '17 at 22:32
• I had no idea, thanks! The integral turns out to be $log(x)$ evaluated from 1 to $\infty$, so I think it's not possible to find the variance in this case. Thanks a lot. – Manuel Jun 4 '17 at 22:37
• This is a special case of the Pareto distribution., which has density of the form $kx^{-(\alpha+1)}$ and for which only moments strictly lower than $\alpha$ are finite. – Glen_b Jun 5 '17 at 0:41