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Let $B,C$ independent random variables such that $B\sim \operatorname{exp}(\lambda),C\sim U[0,1]$.


I have 2 questions about the solution:

  1. "We're looking for the probability that $\mathbb{P}(4B^2-4C>0)$". Why does the coefficient of $B^2$ is $4$ and not $1$? Maybe it's a mistake?
  2. Why does this equality hold? $$ \\ f_{B^2,C}(t,s)=f_{B^2|C}(t|s) \ $$
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    $\begingroup$ 1 is a mistake, no question. $\endgroup$ – Matt Samuel Jan 1 at 17:19
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The first point is a mistake

If $s \in [0,1]$, we have $f_C(s)=1$ since $C$ follows $U[0,1]$.

Hence, when $s \in [0,1]$, $$f_{B^2, C}(t,s) = f_{B^2|C}(t|S)f_C(s)=f_{B^2|C}(t|S)$$

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