# Probability of having real roots

Let $U,V,W$ are independent random variables with $\mathrm{Uniform}(0,1)$ distribution. I am trying to find the probability that $Ux^{2}+Vx+W$ has real roots, that is, $P(V^{2}-4UW> 0)$ I have solved this question using double integral but how to do this using triple integral. My Approach: I started with cdf: $P(V^{2}-4UW >0) =P(V^{2} > 4UW) = P(V>2\sqrt{UW})$ = $\int\int_{2\sqrt{uw}}^1 P(V>2\sqrt{UW}) dU dW$ =$\int\int\int_{2\sqrt{uw}}^1 vdU dW dV$

I am finding hard time to get the limits of integral over the region in 3 dimensions.

Using double integral: $P(V^{2}-4UW >0) =P(V^{2} > 4UW) = P(-2\ln V <-\ln 4 - \ln U - \ln W) = P(X <-\ln 4 +Y)$ where $X=-2 \ln V, Y = - \ln U -\ln W$ $X$ has $\exp(1)$ and $Y$ has $\mathrm{gamma}(2,1)$ distribution. $P(X <-\ln 4 +Y) = \int_{\ln4}^\infty P(X < -\ln 4 +Y) f_Y(y) dy$ $$=\int_{\ln 4}^\infty\int_0^{-\ln 4+y} \frac{1}{2} e^{-\frac{x}{2}}ye^{-y} dxdy$$ Solving this I got $0.2545$.

Thanks!

• Your integrals make no sense, because $U$ and $u$ have different meanings (one is a random variable, the other is a dummy variable of integration) and $W$ and $w$ have different meanings. What solution did you get using double integrals and why is it important to obtain one using triple integration? (I obtain $5/36 + \log(2)/6 \approx 0.254413$.) – whuber Oct 14 '13 at 18:37
• @whuber Using double integral, I obtained 0.2545. I want to see how can we solve using triple integral. – user30438 Oct 14 '13 at 18:51
• Perhaps you should show your demonstration using double integrals, because your triple integrals still make no sense. – whuber Oct 14 '13 at 18:57
• @whuber I have posted the solution using double integral. – user30438 Oct 14 '13 at 19:14

Here is a solution without multiple integrals calculation (because I don't like multiple integrals). Actually it only uses three elementary simple integrals. $$P(V^{2}-4UW \leq 0) = E\bigl[P(V^{2}-4UW \leq 0 \mid U,W)\bigr] = E\bigl[f(U,W)\bigr]$$ where $f(u,w)=P(V^{2}-4uw \leq 0)= \min\bigl\{1, 2\sqrt{uw}\bigr\}$. $$E\bigl[f(U,W)\bigr] = E[g(W)]$$ where \begin{align} g(w) & = E\bigl[\min\bigl\{1, 2\sqrt{Uw}\bigr\}\bigr] = 1 \times \Pr(2\sqrt{Uw}>1) + E\bigl[2\sqrt{Uw} \mathbf{1}_{2\sqrt{Uw}\leq 1}\bigr] \\ & = \Pr(U>\frac{1}{4w}) + 2\sqrt{w}E\bigl[\sqrt{U} \mathbf{1}_{U \leq \frac{1}{4w}}\bigr] \\ & = \max\bigl\{0, 1 - \frac{1}{4w}\bigr\} + 2\sqrt{w} \times \frac{2}{3} \times \min\bigl\{1, \frac{1}{{(4w)}^{\frac{3}{2}}}\bigr\} \\ & =\begin{cases} 0 + \frac{4}{3}\sqrt{w} & \text{if } w \leq \frac{1}{4} \\ 1 - \frac{1}{4w} + \frac{1}{6w} & \text{if } w > \frac{1}{4} \end{cases}, \end{align} and we get $$E[g(W)] = \frac{1}{9} + \frac{3}{4} - \frac{1}{12} \log 4 = \frac{31}{36}-\frac{\log 2}{6},$$ and finally $$P(V^{2}-4UW > 0) = \frac{5}{36} + \frac{\log 2}{6} \approx 0.2544134.$$
• @whuber Why mysterious ? I only use the formula $E[f(X,Y)]=E[E[f(X,Y) \mid Y]]$ at each step. – Stéphane Laurent Oct 14 '13 at 21:18
• The mysteries lie primarily in the details, which are carried out wholly without explanation. A look at the question shows the O.P.'s effort fell apart in not recognizing the importance of tracking the domains of the RVs, as evidenced by the appearances of "$\min$" and "$\max$" in this answer. Thus, although you have provided a correct answer, it does little to reveal what went wrong or to explain the methods you have used to break up the integrals and identify the proper ranges to use in each one. – whuber Oct 14 '13 at 21:24
$P(V^{2} - UW \geq 0) = P(V^{2} \leq UW, 0\leq U,V,W \leq 1)$ $=P(0 \leq V \leq 2\sqrt{UW},0\leq U,V,W \leq 1)$ $=P(0 \leq V \leq min(1,2\sqrt{UW}),0\leq U,V,W \leq 1)$ $=\int_0^\frac{1}{4}\int_0^1\int_0^{2\sqrt{uw}}dudvdw + \int_\frac{1}{4}^1\int_0^{\frac{1}{4u}}\int_0^{2\sqrt{uw}}dudvdw + \int_{\frac{1}{4}}^1\int_\frac{1}{4u}^1\int_0^1dudvdw$