I know that: \begin{equation}\displaystyle P(A=x|A+B=y) = \frac{P(A=x \cap A+B=y)}{P(A+B=y)}\end{equation} Assuming $A$ and $B$ are independent, the intersection of the two events can be resolved as follows: \begin{equation} P(A=x \cap A+B=y)=P(A=x)\cdot P(x+B=y) \end{equation} I don't know how to follow the same steps for the conditional probability with inequality: \begin{equation}P(A<B | C<B)\end{equation} where A, B and C are independent random variables.



If we can assume $A,B,C$ are continuous random variables, then,

\begin{eqnarray*} P(A \lt B \mid C \lt B) &=& \dfrac{P(A \lt B \;\cap\; C \lt B)}{P(C \lt B)}. \end{eqnarray*}

$$P(C \lt B) = \int_{-\infty}^\infty f_B(x)F_C(x)\,dx.$$

\begin{eqnarray*} P(A \lt B \;\cap\; C \lt B) &=& \int_{-\infty}^\infty f_B(x)F_{AC}(x,x)\,dx \\ &=& \int_{-\infty}^\infty f_B(x)F_A(x)F_C(x)\,dx \quad\mbox{(by independence)}.\end{eqnarray*}

Applying this to a simple example where $A,B,C$ each have uniform distribution over $[0,1]$:

$$P(C \lt B) = \int_0^1 1\cdot x\,dx = \left[\frac{x^2}{2} \right]_0^1 = \frac{1}{2}.$$

$$P(A \lt B \cap C \lt B) = \int_0^1 1\cdot x\cdot x\,dx = \left[\frac{x^3}{3} \right]_0^1 = \frac{1}{3}.$$

$$\therefore P(A \lt B \mid C \lt B) = \dfrac{1/3}{1/2} = \dfrac{2}{3}.$$

  • $\begingroup$ Thanks @Mick A. Do you know a textbook where I can find further explanation in this specific area? All those I saw explain this part superficially. $\endgroup$ – Albert Sep 29 '14 at 19:51
  • $\begingroup$ @Albert Probability and Random Processes by Grimmett and Stirzaker is my preferred text for probability theory. Chap 4 discusses the above area on continuous RVs and their distribution and density functions. It's quite a thorough book but still easy-reading. $\endgroup$ – Mick A Sep 30 '14 at 10:32

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