$U$, $V$ are two independent random variables each with the uniform distribution on $[0,1]$. Show that $P(V^2>U>x)$ is $1/3 -x +2/3x^{2/3}$ for $0<x<1$. I don't know how to go about showing that $P(V^2>U)$.
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Here is a hint: Draw a square in the Cartesian coordinate plane with vertices $$\{(0,0), (1,0), (1,1), (0,1)\}.$$ Label the horizontal axis $U$, the vertical axis $V$. Now draw a vertical line through the square, and call its intersection with the horizontal axis the value $x$. Shade the region inside the square to the right of this line. The points in this region represent the $(U,V)$ pairs that satisfy $U > x$.
Next, draw the parabola $V^2 = U$, which will have its vertex at $(U,V) = (0,0)$, a point through $(U,V) = (1,1)$, and another point through $(U,V) = (1/4, 1/2)$. In fact, you can plot any number of such points on the parabola; e.g., $(1/9, 1/3)$, and $(4/9, 2/3)$. Shade the region above this curve inside the square. This is the region of $(U,V)$ pairs that satisfy $V^2 > U$.
Now the intersection of these two regions that you shaded is the set of all $(U,V)$ pairs that satisfy the combined inequality $$V^2 > U > x$$ for the given vertical line you drew. How do you find the area of this region as a function of $x$? This is just an exercise in integration.
heropup
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Because: $$\begin{align} \mathsf P(V^2>U>x) & = \int_x^1 P(V^2>u\mid U=u)f_U(u)\operatorname d u \\ & = \int_x^1 P(V^2>u)\,f_U(u)\operatorname d u & \text{by independence} \\ & = \int_x^1\int_{\sqrt{u}}^1 f_V(v)\;f_U(u)\operatorname d v\operatorname d u \end{align}$$
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Graham Kemp
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