Thread: Probability of a quadratic equation to have a solution

Hi,

Let ksi, eta and dzeta to be iid rv ~ Uniform(0,1). What is the probability that the equation: ksi*x^2 + eta*x + dzeta = 0, to have a real solution i.e. D>=0. I got that I need to specify the limits of the tripple integral I have to solve for finding this probability... The point is that I don't know how to solve: (eta^2)/4>=ksi*dzeta. The outer integral would be from 0 to 1, but what about the inner two?

Some help will be much appreciated. Thanks.

This is just a nasty calc problem.
You want to integrate the density of 1, over the region

but only in the unit cube.

The best way to start is to find the boundary, when

and then try to make sense of your region, or the complementary region.

Originally Posted by dely84

Hi,

Let ksi, eta and dzeta to be iid rv ~ Uniform(0,1). What is the probability that the equation: ksi*x^2 + eta*x + dzeta = 0, to have a real solution i.e. D>=0. I got that I need to specify the limits of the tripple integral I have to solve for finding this probability... The point is that I don't know how to solve: (eta^2)/4>=ksi*dzeta. The outer integral would be from 0 to 1, but what about the inner two?

Some help will be much appreciated. Thanks.

This is quite a nice problem for demonstrating hit-and-miss Monte-Carlo. We generate a large sample of points in the unit cube count number of cases where discriminant is positive and divide by the sample size.

Code:

>NN=10000000;       ..sample size
>
>xx=random(3,NN);   ..sample of size NN of coefficients
>
>pp=sum(xx(1,: )^2>4*xx(2,: )*xx(3,: ))/NN ..count the positive disc and divide by sample size
     0.254644 
>
>
>SE=sqrt((NN*pp*(1-pp)))/NN  ..approx standard error of the estimate
     0.000137768 
>

So the required probability is ~=0.255 or 25.5%

CB

Originally Posted by matheagle

This is just a nasty calc problem.
You want to integrate the density of 1, over the region

but only in the unit cube.

The best way to start is to find the boundary, when

and then try to make sense of your region, or the complementary region.

Hi,
Yeah, that is exactly the issue. I am not sure how to do that. Could you please elaborate on how to define the boundaries of the integration over the triple integral. I could solve it just if new how to setup the last boundary.

There's actually another route you can go with this. You can make use of the following fact:

If X ~ uniform[0, 1] then -log(X) ~ exponential(1).

After an annoying amount of calculus and some standard results from distribution theory, the answer I get is:

Originally Posted by Guy

There's actually another route you can go with this. You can make use of the following fact:

If X ~ uniform[0, 1] then -log(X) ~ exponential(1).

After an annoying amount of calculus and some standard results from distribution theory, the answer I get is:

Hi, Thanks for the answer. Could you please show the steps for this transformation. I need to see how to form the integral. If you could post it and I will sove it for me.

Originally Posted by dely84

Hi, Thanks for the answer. Could you please show the steps for this transformation. I need to see how to form the integral. If you could post it and I will sove it for me.

Well, if x, y, z ~ iid uniform[0, 1] then



Now, -2log(y) ~ exponential(2) [I'm going to use the scale parametrizations of the exponential and gamma], and because x and z are independent, -log(x) - log(z) is the sum of two independent exponential(1) random variables, and so is distributed gamma(2, 1). So, if A ~ exponential(2) and B ~ gamma(2, 1) with A independent of B you want to calculate



It should be obvious now how to proceed. If you don't know from class the result about -log(x) giving an exponential, it is really very easy to show using the so-called distribution function method:



which is the cdf of an exponential random variable.