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Given that $x$, $b_1$ and $b_2$ are uniformly distributed $U~(0, 1)$ and $b_1 \leq x$ and $b_2$ is independent. Taking reference from a previous post. Is the following method of calculating $\mathbb{P} ( x \geq b_1 \geq b_2)$ right?

\begin{align*} \mathbb{P} ( x \geq b_1 \geq b_2) &= \int_0^1\int_{b_2}^1\int^1_{b_1} dxdb_1d_2 \end{align*}

albusSimba
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1 Answers1

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If you already know that $x \geq b_1$ then $P(x \geq b_1 \geq b_2)=P(b_1 \geq b_2)=\int_0^{1} \int_0^{b_1} f_{b_2} (y)dy f_{b_1}(t) dt$ assuming that $b_1$ and $b_2$ also take values in $(0,1)$. Make appropriate changes if their ranges are different.

Kavi Rama Murthy
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