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Lets say I have two Exponential $X$ and $Y$ independent variables distributed $Exp(1)$

  • The CDF $F(x)=P(X \leq x)$ is given by $1-e^{-x}$
  • The Survival function $1-F(x) = S(x)=P(X > x)$ is given by $e^{-x}$

Now I want to know the intersection and union of some events $A_x$ and $A_y$ lets say that I believe that they occured between $[0, 1]$ and they did not occur between $(1,\infty]$. I would like to calculate the probability that some events did occur and some didn't. Am I thinking this correctly?

$P(A_x \cap \overline A_y)$ = $P(A_x) \cdot P(\overline A_y)$ = $P(X \leq 1) \cdot P(Y > 1) = (1-e^{-1})(e^{-1}) = e^{-1}-e^{-2}$

$P(A_x \cup A_y) = P(A_x) + P(A_y) - P(A_x \cap A_y) = P(X \leq 1) + P(Y \leq 1) - P(X \leq 1) \cdot P(Y \leq 1) = 2e^{-1} - e^{-2}$

Or do I have to make a joint PDF of $X$ and $Y$ and then calculate the probabilities?

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No, due to the independence, you don't have to explicitly use the joint pdf. Your calculations are correct, but needs some improvements:

  • Event notation is not good. $X$ and $Y$ are random variables, but you also use them to denote the events $X\leq 1$ and $Y\leq 1$. Letting them to be $A_x, A_y$ respectively is much less confusing.

  • The formula for $P(A_x\cup A_y)$ is actually $P(A_x)+P(A_y)-P(A_x\cap A_y)$ (the last term is the intersection, not union), but you've already done the final calculation correct, although the first term in the next expression seems to be multiplication but executed as addition (as it's supposed to be).

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  • $\begingroup$ Thank you for your answer and input. You're right and I've mixed them up together. I will edit it $\endgroup$
    – arezaie
    Mar 20, 2021 at 13:44
  • $\begingroup$ Also regarding $ P(A_x \cap A_y)$ and the multiplication/addition error that was a typo. I fixed it now :) $\endgroup$
    – arezaie
    Mar 20, 2021 at 13:50

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