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.
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.
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 $\displaystyle \frac{1}{3} - x + \frac{2}{3} x^{\color{red}{3}}$ ....