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Let X and Y be independent continuous random variables with marginal CDFs given by

$F_X(x) =\begin{cases} 0 & \text{for }x < 0\\ x/3 & \text{for }0 \leq x \leq 3\\ 1 & \text{for }x > 3 \end{cases}$

$F_Y(y) = \begin{cases} 0 & \text{for }y < 0\\ y^2/16 & \text{for } 0 \leq y \leq 4\\ 1 &\text{for } y > 4\end {cases}$

Find $P(X \leq 2, Y \leq 1\mid X \leq 1, Y \leq 3)$.

I began by making a joint CDF. Do I just use the Bayes' Rule using the CDF?

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If they are independent then try $$P(X \le 2, Y \le 1|X \le 1, Y \le 3) = P(X \le 2|X \le 1)P(Y \le 1|Y \le 3)$$

$P(X \le 2|X \le 1)$ is easy.

$P(Y \le 1|Y \le 3)=\dfrac{P(Y \le 1)}{P(Y \le 3)}.$

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  • $\begingroup$ Isn't $ P (X \leq 2 |X \leq 1) $ simply just 1? $\endgroup$ – Ufuk Can Bicici Jul 24 '14 at 21:34
  • $\begingroup$ @Ufuk Can Biçici: I did say "easy" $\endgroup$ – Henry Jul 24 '14 at 21:59
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$ P ( X \leq 2, Y \leq 1 |X \leq 1, Y \leq 3) $ is equal to $ \dfrac {P (X \leq 2, Y \leq 1 ,X \leq 1, Y \leq 3)}{P (X \leq 1, Y \leq 3)}$. We can take the intersections of the intervals which are belonging to the same random variables and this makes a reduction to $ \dfrac {P (X \leq 1, Y \leq 1)}{P (X \leq 1, Y \leq 3)} $ Since both random variables are independent this reduces to$ \dfrac {P (X \leq 1) P(Y \leq 1)}{P (X \leq 1) P(Y \leq 3)} $. So you only need to calculate $\dfrac {P(Y \leq 1)}{P(Y \leq 3)} $ in the end. This means just calculate $\dfrac {F_Y (1)}{F_Y (3)}$ and you are over. I don't think that Bayes rule is necessary here.

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