In the following second order equation $ax^2+2bx+1.5=0$ where $a$ and $b$ are given by random points $(a,b)$ in the $[0,2]\times[0,1]$ rectangle, what is the probability of having two real solutions?

I'm a little lost here. I tried integrating $4b^2-6a$ with $a=0\to 2$ and $b=0\to 1$ as limits but the integral comes up negative. I created a simulation of the problem using matlab and the probability is 0.11 but I want to find a way to solve it on paper and not with using matlab.

Any thoughts?

  • $\begingroup$ Hint: I think you are confusing taking the expectation of a certain quantity with finding the probability over a particular region of interest bounded by a function. $\endgroup$ – cardinal Apr 1 '12 at 19:24
  • $\begingroup$ Should your equation be $a x^2 + 2b + 3/2 = 0$ or $a x^2 + 2 b x + 3/2 = 0$? It seems the latter, but the former is what is given. $\endgroup$ – cardinal Apr 1 '12 at 19:30
  • $\begingroup$ it's the latter, i missed an x by mistake. $\endgroup$ – System Apr 1 '12 at 19:38
  • $\begingroup$ Crossposted here: math.stackexchange.com/questions/126985/… $\endgroup$ – cardinal Apr 1 '12 at 21:42

The quadratic formula tells us this equation has two real solutions exactly when the discriminant $4b^2 - 6a$ is positive. This describes a set of points $R$ in the $(a,b)$ plane, shown in blue here:


When the joint density of $(a,b)$ is $f$, then (by definition of pdf) the probability of any event (like $R$) is given by its integral $\int_R f(a,b)da\ db$. Because the distribution is uniform, this is the same as finding the area of the shaded region as a proportion of the total area (equal to $2$), often written as a double integral like

$$\frac{1}{2}\int_0^1 \int_{a\lt 4b^2/6,\ 0\le a\le 2} da\ db.$$

However, we can reduce this to a single integral: the shaded region is bounded by the parabola $a = 2 b^2/3$ on the right, $a=0$ on the left, and extends from $b=0$ to $b=1$. Its area therefore is $\int_0^1 2b^2/3\ db = 2/9$. That amounts to $1/9 = 0.1111\ldots$ of the total area.

Edit (see comments). In case this is unclear, we can proceed more formally. The uniform distribution function $f$ is obtained by knowing (a) it is constant on the rectangle $[0,2]\times[0,1]$ and (b) is zero outside this set. From (a) and the fact that any PDF must integrate to unity forces $f(a,b)=1/2$ inside the rectangle, whence

$$\eqalign{ f(a,b) = &1/2, &0 \le a \le 2, 0 \le b \le 1\\ &0 &\text{otherwise}. }$$

Integration is defined in terms of characteristic functions: the integral over an event $E$ with respect to a measure $d\mu$, written, $\int \cdots \int_E d\mu$, equals $\int \cdots \int I_E(x) d\mu(x)$ where the multiple integral is taken over all possible values of $x$ and $I_E(x) = 1$ when $x\in E$ and $I_E(x)=0$ otherwise. The figure immediately shows that the solution is

$${\Pr}_f[(a,b)\in R] = \int \int_R f(a,b)\ da \ db$$

and the original double integral expression follows immediately from this expression by the definition of integration. For more about this, consult any textbook on measure theory and integration or--for a less formal approach--consult any advanced calculus text that covers multiple integration. End of edit.

This is essentially problem #50 from Fred Mosteller's Fifty Challenging Problems in Probability:

What is the probability that the quadratic equation $x^2 + 2bx + c = 0$ has real roots?

Solving it requires proposing some "reasonable" probability distribution for $(b,c)$. Mosteller chooses a set of uniform distributions over a sequence of rectangles that grows without bound and takes the limit.

  • $\begingroup$ (+1) I always like how attractive the graphics you make are and how efficient you are at creating them. :) $\endgroup$ – cardinal Apr 1 '12 at 19:32
  • 1
    $\begingroup$ Thank you, cardinal. As you probably know, the credit goes to Mathematica. Here, the plot is produced by the command RegionPlot[4 b^2 - 6 a > 0, {a, 0, 2}, {b, 0, 1}, AspectRatio -> 1/2] $\endgroup$ – whuber Apr 1 '12 at 19:35
  • $\begingroup$ The tool can be powerful, but it still takes the craftsman's hand. $\endgroup$ – cardinal Apr 1 '12 at 19:36
  • $\begingroup$ @whuber the integral is 4 b^2 - 6 a ? $\endgroup$ – System Apr 1 '12 at 20:14
  • $\begingroup$ System, as explained below the figure in my reply, the integral is $\int_0^1 2b^2/3\ db$. I will edit the reply to make this clearer. $\endgroup$ – whuber Apr 1 '12 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.