Chapter 5

Question 31: Let U be a uniformly distributed random variable on [0, 1]. What is the probability that the equation

x2 + 4Ux + 1 = 0

\[x^2 + 4Ux + 1 \]

has two distinct real roots x1 and x2?

Solution:

To find the probability that the equation has two distinct real roots, we can analyze the discriminant of the quadratic equation.

namely if D>0 then we know that the quadratic has two distinct real roots

Given that \[D = b^2 -4ac\]

\[D = (4U)^2 -4(1)(1) > 0\]

\[D = 16U^2 -4 > 0\]

\[D = 4U^2 -1 > 0\]

\[D = (2U + 1)(2U - 1) > 0\]

we can say that the quadratic will have two real roots if U is in the range (12,1]. Since this is exactly half the total range of U, the probability of the quadratic having two distinct real roots is 0.5.

We can confirm this using the following R code:

set.seed(seed = 1234)

simulation <- 10000
Us <- runif(simulation)
num_real_distinct <- 0

for (U in Us){
  poly_coef <- c(1, 4 * U, 1)
  roots <- polyroot(poly_coef)
  check_z <- round(Im(roots[1]), 10) == 0 & round(Im(roots[1]), 10) == 0
  check_distinct <- roots[1] != roots[2]
  if (check_z  & check_distinct){
    num_real_distinct <- num_real_distinct + 1
  }
}

table(Us > 0.5)
## 
## FALSE  TRUE 
##  4990  5010

We get that the quadratic had two distinct, real solutions 50.1% of the time.

print(num_real_distinct/10000)
## [1] 0.501