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Suppose that $A, B,$ and $C$ are independent random variables, each being uniformly distributed over $(0,1)$. What is the probability that $AX^2 + BX + C$ has real roots?

I am given a hint that if $X$ ~ uniform $(0, 1)$, then $-\ln(X)$ is exponential. The sum of two (or more) independent exponential random variables is gamma...

How does this hint help answer this question?

Thanks a lot

afsdf dfsaf
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  • we need $b^2-4ac$ to be greater than 0.Now take log on both sides you get $ln(b)$ on one side and sum of two ln on other side – happymath Mar 02 '14 at 06:22
  • could you write it out..."take log on both sides.....two ln on other side" was not so clear to me – afsdf dfsaf Mar 02 '14 at 06:25
  • Calculate the @g76989b integral. – Felix Marin Mar 02 '14 at 06:26
  • I am told the following: $$\begin{align} \int_0^1 \int_0^1 \int_{\min{1, \sqrt{4ac}}}^1 1 ;\text{d}b,\text{d}c,\text{d} &a= \int_0^1 \int_0^{\min{1, 1/4a}}\int_{\sqrt{4ac}}^1 1;\text{d}b,\text{d}c,\text{d}a\ &= \int_0^{1/4} \int_0^1 \int_{\sqrt{4ac}}^1 1;\text{d}b,\text{d}c,\text{d}a + \int_{1/4}^1 \int_0^{1/4a}\int_{\sqrt{4ac}}^1 1;\text{d}b,\text{d}c,\text{d}a \end{align}$$ why the middle integrate from 0 to min{1, 1/4a} from the second integral.>.where does 1/4a come from? why the min{...} does not go to the front integral? why they break up into last step like this – afsdf dfsaf Mar 02 '14 at 06:42

1 Answers1

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Taking Logarithm yields

$2*ln(b)=ln(4)+ln(a)+ln(c)$. On the LHS we have an exponential random variable on the RHS we have sum of two exponential random variables which is Gamma.

On the right it is $\Gamma(2,1)$ on the left side it is $\Gamma(1,1)$.

happymath
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  • could you go on a little further...as I still do not see how to approach it? – afsdf dfsaf Mar 02 '14 at 06:31
  • first of all, you have to move the equation into $-ln(4) - ln(a) - ln(c) = -2ln(b)$...then my question why $\Gamma(2,1)$ and $\Gamma(1,1)$? – afsdf dfsaf Mar 02 '14 at 06:40
  • @afsdfdfsaf in the pdf of $\Gamma$ put 1 and 1 for the 2 parameters it will become exponential – happymath Mar 02 '14 at 06:43
  • you lost me again...could you explain a little bit more? – afsdf dfsaf Mar 02 '14 at 06:44
  • @afsdfdfsaf pdf of Gamma distribution is $1/(\Gamma(k)\theta^k)$$x^{k-1}$$e^{-x/\theta}$. Put k=1 and $\theta$ =1 – happymath Mar 02 '14 at 06:48
  • I got put k=1 and $\theta =1$ into gamma distribution will give us exponential...how does $\Gamma(2,1)$ and $\Gamma(1,1)$ help here? What are the roles for them? – afsdf dfsaf Mar 02 '14 at 06:53
  • By the way, what does the two coordinates in $\Gamma(2,1)$ is associated with since for Gamma pdf, we only have $\Gamma(\text [ $one$ $variable$])$? – afsdf dfsaf Mar 02 '14 at 07:10