Find Mean/Variance from probability mass function

**komissar** · September 12th 2009, 10:22 PM

I'm not sure if this actually Advanced, but it is from a 3000-level University course. My apologies if I am misposting.

A random variable X has the probability mass function
$f(x) = p * 0.6^{x-1}$ , for x = 1, 2 , 3, …
Find p, the mean and the variance of X.

Ok... So that seems pretty clear to me that p = .4 (based on the logic that q=.6 = 1-p).

But I have no idea where to go from here. If I go to take the mean and variance, according to the text and notes, the formulae I'd be using are:

$ \mu = \sum x f(x) = \sum_{x=1}^{\infty} x (.4 * .6^{x-1}) $

and

$ \delta^2 = \sum(x-\mu)^2f(x) $

Maybe my calculus-fu has weakened (it's been a long time), but I can't for the life of me figure out how to do those sums since they don't appear to resemble any of the series I studied in Calc II...

If anyone could give me a pointer, or tell me where I did something wrong, I'd appreciate it. Thanks...

**CaptainBlack** · September 12th 2009, 11:15 PM

Originally Posted by komissar

I'm not sure if this actually Advanced, but it is from a 3000-level University course. My apologies if I am misposting.

Ok... So that seems pretty clear to me that p = .4 (based on the logic that q=.6 = 1-p).

That is the right value for $p$ , but your explanation cannot be followed. You should state that for f to be a pmf:

$\sum_{x=1}^{\infty} p (0.6)^{x-1}=1$

or state some other obvious reason why this is so.

CB

**CaptainBlack** · September 12th 2009, 11:19 PM

Originally Posted by komissar

But I have no idea where to go from here. If I go to take the mean and variance, according to the text and notes, the formulae I'd be using are:

$ \mu = \sum x f(x) = \sum_{x=1}^{\infty} x (.4 * .6^{x-1}) $

Consider:

$g(x)=0.4 \; \sum_{x=1}^{\infty} 0.6^{x}$

Now differentiate $g(x)$ (and the series term by term)

CB

**CaptainBlack** · September 12th 2009, 11:20 PM

Originally Posted by komissar

...
and

$ \sigma^2 = \sum(x-\mu)^2f(x) $

Use:

$\sigma^2= \overline{x^2}-\overline{x}^2 = \overline{x^2} - \mu^2$

CB

**komissar** · September 12th 2009, 11:27 PM

Originally Posted by CaptainBlack

That is the right value for $p$ , but your explanation cannot be followed. You should state that for f to be a pmf:

$\sum_{x=1}^{\infty} p (0.6)^{x-1}=1$

or state some other obvious reason why this is so.

I was just putting that there to illustrate my thought process in case that's where my error was, but I'll keep it in mind. Thank you.

Originally Posted by CaptainBlack

Consider:

$g(x)=0.4 \; \sum_{x=1}^{\infty} 0.6^{x}$

Now differentiate $g(x)$ (and the series term by term)

I'm a little lost there. Is that a rewriting of the function I had, or was mine wrong?

Originally Posted by CaptainBlack

Use:

$\sigma^2= \overline{x^2}-\overline{x}^2 = \overline{x^2} - \mu^2$

CB

I assume it'll be summed the same way you described for the mean above?

Thanks again.

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