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Let $U$ and $V$ be independent continuous random variables, identically distributed uniformly over $[0,1]$. Show that for $0 \leq x\leq1$ , $$P(x < V < U^2)= \frac{1}{3} - x + \frac{2}{3} x^{3/2}$$ and hence write down the density of $V$ conditional on the event $V< U^2$.

How should I introduce the conditional probability aspect into the solution ? Can this be rewritten as the cumulative distribution of joint variables ?

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  • $\begingroup$ So, are you having a problem with the first part? The rest of the question seems to indicate your only problem is converting that result to a statement about conditional probabilities. $\endgroup$ – Thomas Andrews Sep 10 '14 at 17:29
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Here is a hint. Draw a square in a Cartesian coordinate plane with vertices $(0,0), (1,0), (1,1), (0,1)$. Label the horizontal axis $V$, and the vertical axis $U$. Draw a vertical line $V = x$ for some $x \in [0,1]$. Draw the curve $V = U^2$, which is a parabola with vertex at $(0,0)$ and axis of symmetry coinciding with the line $U = 0$. Now consider the area of the region inside the square that satisfies $x < V < U^2$ as a function of $x$.

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  • $\begingroup$ Great idea ! But how do I justify that the area represents a probability ? $\endgroup$ – user175081 Sep 10 '14 at 17:38
  • $\begingroup$ $$\Pr[x < V < U^2] = \int_{v=x}^1 \int_{u=\sqrt{v}}^1 f_U(u) f_V(v) \, du \, dv.$$ $\endgroup$ – heropup Sep 10 '14 at 19:28
  • $\begingroup$ @user175081 It's because $U$ and $V$ are each independently and uniformly distributed over unit lengths. $\endgroup$ – Graham Kemp Sep 10 '14 at 23:08
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Assuming you've proved the first part.

Using: $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$.

Then $P(x<V\mid V<U^2) = \frac{P(x<V<U^2)}{P(V<U^2)}$. Then:

$$P(V\leq x\mid V<U^2)=1-P(x<V\mid V<U^2)$$ is the CDF of $V$ given that $V<U^2$.

So the derivative of the CDF would be the PDF of $V$ given $V<U^2$.

So $P(V<U^2)=P(0<V<U^2)=\frac{1}{3}$ and $P(x<V<U^2)=\frac{1}{3}-x+\frac{2}{3}x^{3/2}$. The rest is just arithmetic.

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