1. ## Uniform variate problem

Let X be a uniform variate defined on (-k,k). Determine k so that
P(|X|<1) = P(|X|>2)

Let X be a uniform variate defined on (-k,k). Determine k so that
P(|X|<1) = P(|X|>2)

3

3. ## Could u please tell me how u got the answer

Originally Posted by matheagle
3

How did u arise to this conclusion???

4. Hello,

Draw a line where you represent -2,-1,0,1,2 and represent the areas where |X|<1 and |X|>2 and you'll see that for the areas to be equal, the domain has to stop to -3 and 3.

More formally... :

$\displaystyle P(|X|<1)=\frac{1}{2k} \int_{-1}^1 ~dt=\frac 1k$

$\displaystyle P(|X|>2)=\frac{1}{2k}\left(\int_{-k}^{-2} ~dt+\int_2^k ~dt\right)=\frac{1}{2k}\cdot (-2+k+k-2)=\frac{k-2}{k}$

Thus we must have $\displaystyle \frac 1k=\frac{k-2}{k} \Rightarrow 1=k-2\Rightarrow k=3$

boldeagle : what's the point giving the solution without further information ? It doesn't help

5. Fine, 1+2=3, since....

$\displaystyle 2\int_0^a{dx\over 2k}=2\int_b^k{dx\over 2k}$

cancelling the constants leaves $\displaystyle \int_0^adx=\int_b^kdx$

Thus a=k-b hence k=a+b.