Have the following continuous random variable....

**breitling** · January 13th 2011, 10:09 AM

Hi I know this is a relatively easy question but im a bit stuck, previous one i have calculated i have found the cumulative function and plug in the upper limit and equal it to 1 and from the find the constant. Now with this having no upper limit im not to sure. The rest of the question i know how to calculate, but you need to find the constant in order to work it out, the question is as follows:
f(x)=c/(x^3) for x greater or equal to 1
0 otherwise.

1) find the value of the constant c.

Many thanks in advance.

**mr fantastic** · January 13th 2011, 10:14 AM

Originally Posted by breitling

Hi I know this is a relatively easy question but im a bit stuck, previous one i have calculated i have found the cumulative function and plug in the upper limit and equal it to 1 and from the find the constant. Now with this having no upper limit im not to sure. The rest of the question i know how to calculate, but you need to find the constant in order to work it out, the question is as follows:
f(x)=c/(x^3) for x greater or equal to 1
0 otherwise.

1) find the value of the constant c.

Many thanks in advance.

"for x greater or equal to 1" means $1 \leq x < + \infty$ .

**breitling** · January 13th 2011, 10:26 AM

Originally Posted by mr fantastic

"for x greater or equal to 1" means $1 \leq x < + \infty$ .

Exactly what i thought, from this would you find the continuous function, then use x=1?

**harish21** · January 13th 2011, 10:58 AM

Remember the properties of a probability density function..

and since the random variable is continuous,

$\displaystyle \int_1^{\infty} \dfrac{c}{x^3}\;dx\;=\;1$

find c

**mr fantastic** · January 13th 2011, 12:54 PM

Originally Posted by harish21

Remember the properties of a probability density function..

and since the random variable is continuous,

$\displaystyle \int_1^{\infty} \dfrac{c}{x^3}\;dx\;=\;1$

find c

Yes, and since it's an improper integral care must be taken in evaluating it (using limits etc.).

Originally Posted by breitling

Exactly what i thought, from this would you find the continuous function, then use x=1?

You ought to have examples in your textbook or class notes on this type of situation.

**breitling** · January 14th 2011, 05:09 AM

Originally Posted by mr fantastic

Yes, and since it's an improper integral care must be taken in evaluating it (using limits etc.).

You ought to have examples in your textbook or class notes on this type of situation.

Unfortunately to my surprise i havent, we only have them them for the kind i have said find the cumulative and then use the upper limit.
So honestly haven't got a clue, any help would be most appreciated.

**breitling** · January 14th 2011, 05:33 AM

Would i be correct to think that the integral between infinity and 0 is:
-c/(2x^2)=1
?

**mr fantastic** · January 14th 2011, 01:35 PM

Originally Posted by breitling

Would i be correct to think that the integral between infinity and 0 is:
-c/(2x^2)=1
?

No.

$\displaystyle 1 = \int_1^{+\infty} \frac{c}{x^3} \, dx = \lim_{\alpha \to +\infty} \int_1^{\alpha} \frac{c}{x^3} \, dx = ....$

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