# Thread: random variable X and random variable X^2

1. ## random variable X and random variable X^2

Well this is the question in the tutorial:

A random variable X has the density function f(x) = c/(x^2 +1), x is a real number.
a) Find the value of the constant c (already solved) [ans:1/pi]
b) Find the probability that X^2 lies between 1/3 and 1 [ans: 1/6]

well i understand what they want for part (a) which i had already solved but i don't understand part (b) especially on the X^2. Don't really know how it affects the function given for random variable X. Anyone mind explaining?

2. Hint:

If $\displaystyle 1/3 \leq X^2 \leq 1$, what are the possible values of X?

3. er... sqrt(1/3) < x < sqrt(1)?

4. Hello,

Originally Posted by Danshader
er... sqrt(1/3) < x < sqrt(1)?
Only if x is a positive variable

5. well the question said that x is a real number so it can be both negative and positive.

6. Ok, so :

$\displaystyle x^2 \geq 1/3 \text{ iff } \{ [x \geq \sqrt{1/3} \text{ and } x \geq -\sqrt{1/3}] \Longleftrightarrow x \geq \sqrt{1/3} \}$ OR $\displaystyle \{ [x \leq \sqrt{1/3} \text{ and } x \leq -\sqrt{1/3}] \Longleftrightarrow x \leq -\sqrt{1/3} \}$

The same goes for $\displaystyle x^2 \leq 1$...

If you want to continue this way, you will have 4 probabilities (maybe 3) to calculate...

I dunno if there is an easier way.

7. i just did this:

1/3 < x^2 <1
sqrt(1/3) < x < sqrt(1) and -sqrt(1/3) > x > -sqrt(1)

so this kinda give me two probabilities to solve o.o"

well i solved for the first part and i got 1/12 so since the intervals are of the same range i multiplied the value obtained by 2 to get 1/6 o.o" is
it correct?

8. Wait, I just have a problem :

$\displaystyle x^2 > 1/3$

$\displaystyle x^2-1/3>0$

$\displaystyle (x-\sqrt{1/3})(x+\sqrt{1/3})>0$

So $\displaystyle x-\sqrt{1/3}>0 \ and \ x+\sqrt{1/3}>0$ or $\displaystyle x-\sqrt{1/3}<0 \ and \ x+\sqrt{1/3}<0$

This seems strange to me that the bounds are like this, but I can't find the mistake, and this makes me disagree with your probabilities :/

Edit : nevermind... I was just confused >_<

I didn't check the calculus, but it's correct, since the second inequality is the same as the first one (with x instead of -x) and the density function is even

9. ok i just confused myself >.<.... shouldn't the range be:

sqrt(1/3) < x < 1
-sqrt(1/3) < x < -1

???

10. Look at my edit

To resume :

$\displaystyle P(1/3<X^2<1)=P(\sqrt{1/3}<X<1)+P(-\sqrt{1/3}>X>-1)$ $\displaystyle =P(\sqrt{1/3}<X<1)+P(\sqrt{1/3}<-X<1)=2P(\sqrt{1/3}<X<1)$

11. ok i understand the second line since it is an even function so we can add them up but is the second range of x correct? O.o?

P(-sqrt(1/3) > x > -1)

i think i am gonna read my high school maths back... >.>

12. Originally Posted by Danshader
ok i understand the second line since it is an even function so we can add them up but is the second range of x correct? O.o?

P(-sqrt(1/3) > x > -1)

i think i am gonna read my high school maths back... >.>
Oh it seems correct...
Just multiply all of it by -1 to get the right thing.

13. ok >.> i will take your word for it... or else i am screwed in exam xD

14. Erm... This is too much responsibility for me >.<

But I'm pretty sure it's ok though ^^