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The question, from the textbook: Mathematical Statistics and Data Analysis

Let $U_1, U_2, U_3$ be independent random variables uniform on $[0,1]$. Find the probability that the roots of the quadratic $U_1x^2+U_2x+U_3$ are real.

I know this question has already been asked on StackExchange but I'd like to present my incorrect attempt at it in the hopes that someone could tell me where I went wrong.

So this question boils down to finding $P(U_2^2-4U_1U_3\ge 0)$ which is the discriminant. Which is equivalent to $1-P(-\sqrt{4U_1U_3} \lt U_2\lt \sqrt{4U_1U_3})$

Since all three random variables are uniform on $[0,1]$, their density function would be just be $1$.

Putting it all together, I get the triple integral of

$\int_{0}^{1}\int_{0}^{1}\int_{-\sqrt{4u_1u_3}}^{\sqrt{4u_1u_3}}du_2du_1du_3$

Which is what I think should equal to $P(-\sqrt{4U_1U_3} \lt U_2\lt \sqrt{4U_1U_3})$

The triple integral turns out to be a number greater than 1 which is obviously wrong. Where did I go wrong? Can anything be salvaged here, perhaps a triple integral with different bounds? I saw the solution to this question done by someone else: Probability that a quadratic polynomial with random coefficients has real roots but I don't think I would be able to think of something like that in a test setting. Any pointers would be much appreciated!

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Your errors are that (a) $u_2$ can never be negative, so the lower limit should have been $0$; and (b) your triple integral's innermost upper bound isn't clipped to the unit cube's boundaries. So $u_1$ and $u_3$ are properly confined to be less than $1$, but you have let $u_2$ go as high as $2$ (when $u_1 = u_3 = 1$).

There are a few ways to deal with this second issue. One is to break the expression up into two integrals: one over the domain where $4u_1u_3 \leq 1$ (so, for example, let the middle bounds range from $u_1 = 0$ to $\frac{1}{4u_3}$), over which $u_2$ ranges from $0$ to $\sqrt{4u_1u_3}$; and a second integral over the domain where $4u_1u_3 > 1$ (let the middle bounds range from $u_1 = \frac{1}{4u_3}$ to $1$), over which $u_2$ ranges only between $0$ and $1$.

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  • $\begingroup$ Why are the bounds $u_2$ for the second integral $0$ to $1$. I get that the middle bounds would go from $1/4u_3$ to $1$, but doesn't $4u_1u_3$ being greater than $1$ affect the bounds of $u_2$ in some way? I hope my question makes sense.. $\endgroup$ – PurpleBlueJeans Nov 26 '18 at 2:17
  • $\begingroup$ @PurpleBlueJeans: The upper bound for $u_2$ should be $\min\{1, 4u_1u_3\}$. $\endgroup$ – Brian Tung Nov 26 '18 at 4:50

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