Let U and V be independent and identically distributed uniformly on the interval [0, 1].

Show that for 0<x<1;

P(x<V<U^2)=1/3 - x + (2/3)*x^3

and hence write down the density of V conditional on the event that V<U^2.

- Apr 19th 2009, 09:16 AMdjones90Two independant and identically distributed uniform distributions
Let U and V be independent and identically distributed uniformly on the interval [0, 1].

Show that for 0<x<1;

P(x<V<U^2)=1/3 - x + (2/3)*x^3

and hence write down the density of V conditional on the event that V<U^2.

- Apr 20th 2009, 04:28 AMmr fantastic
1. Note that $\displaystyle f(u, v) = 1$ for $\displaystyle 0 \leq u \leq 1$, $\displaystyle 0 \leq v \leq 1$ and zero otherwise.

2. Draw the square defined by $\displaystyle u = 0, u = 1, v = 0$ and $\displaystyle v = 1$ and shade the region corresponding to $\displaystyle x < v < u^2$.

3. Then $\displaystyle \Pr(x < V < U^2) = \int_{u = \sqrt{x}}^{u = 1} \int_{v = x}^{v = u^2} 1 \, dv \, du$.

I get $\displaystyle \frac{1}{3} - x + \frac{2}{3} x^{\color{red}{3/2}}$, NOT $\displaystyle \frac{1}{3} - x + \frac{2}{3} x^{\color{red}{3}}$ .... - Apr 22nd 2009, 01:49 AMdjones90
my mistake.

Thanks!