Mean of a probability density function

DannyBoy · September 5th 2007, 09:10 AM

I have a probability density function (PDF), $\vartheta(r)$ , where $r\in[0,\infty]$ . So the mean of the PDF ( $\bar{r}$ ) is calculated by $\int_0^\infty r\vartheta(r)dr$ .

One can say that $\lim_{t\to\infty}\frac{\int_0^t r\vartheta(r)dr}{\int_0^t \vartheta(r)dr}=\bar{r}$ .

But is it correct to state that $\frac{\int_0^t r\vartheta(r)dr}{\int_0^t \vartheta(r)dr}\approx\bar{r}$ when $t\gg\bar r$ ?

Thanks in advance!

**TKHunny** · September 5th 2007, 08:00 PM

1) Isn't the denominator of the infinite limit just unity?
2) This is why we have Rules of Thumb. Here are a couple:

Use the t-distribution until the sample size is "big enough".

The Normal is a good approximation of Binomial if n is big enough and p is close to 1/2.

3) I'm not convinced that the mean is a good measure of t being big enough. Outliers are bad!

My views. I welcome others'.

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