# Have the following continuous random variable....

• Jan 13th 2011, 11:09 AM
breitling
Have the following continuous random variable....
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.

• Jan 13th 2011, 11:14 AM
mr fantastic
Quote:

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.

"for x greater or equal to 1" means $1 \leq x < + \infty$.
• Jan 13th 2011, 11:26 AM
breitling
Quote:

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?
• Jan 13th 2011, 11:58 AM
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
• Jan 13th 2011, 01:54 PM
mr fantastic
Quote:

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.).

Quote:

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.
• Jan 14th 2011, 06:09 AM
breitling
Quote:

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.
• Jan 14th 2011, 06:33 AM
breitling
Would i be correct to think that the integral between infinity and 0 is:
-c/(2x^2)=1
?
• Jan 14th 2011, 02:35 PM
mr fantastic
Quote:

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

No.

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