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Using only the axioms of probability, derive the following result using induction:

\begin{align} \Pr\left(\bigcup_{i=1}^\infty A_i\right) = &\sum_{i=1}^\infty P(A_i) - \sum_{i<j}^\infty \Pr(A_i \cap A_j) - \\ &\sum_{i <j<k}^\infty \Pr(A_i \cap A_j \cap A_k) - \cdots - \\ &(-1)^{n+1}\Pr(A_1 \cap A_2 \cap \cdots \cap A_n) \end{align}

I have began the proof by showing that it holds for $n = 2$, and have assumed it holds for $n \le k$ for some arbitrary $k$. Now it just comes to showing that it holds for $k + 1$:

\begin{align} \Pr\left(\bigcup_{i=1}^{k+1} A_i\right) &= \Pr\left(A_{k+1} \cup \bigcup_{i=1}^kA_i\right) \\ &= \Pr(A_{k+1}) + \Pr\left(\bigcup_{i=1}^k A_i\right) - \Pr\left(A_{k+1} \cap \bigcup_{i=1}^kA_i\right) \end{align}

where the last term can be written as:

$$ \Pr\left(\bigcup_{i=1}^k(A_i \cap A_{k+1})\right) $$

This is where I am stuck and cannot simplify the expression to get the desired result. I am positive I am on the right track with the "proof", but do not know what result or axiom to use to continue

Any hints to how I can continue will be appreciated.

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  • $\begingroup$ Remember that you want to prove it for $k+1$ assuming the $k$ case is true. Assuming the $k$ case is true, you can expand out $P(\bigcup_{i=1}^k X_i)$ for any collection of $k$ sets $X_1, X_2, \ldots, X_k$. The second and third terms on the RHS of your second equation are unions of $k$ sets, so you can now expand them. $\endgroup$ – Paul Sep 27 '18 at 21:26
  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ – gung Oct 24 '18 at 14:53
  • $\begingroup$ Did @Sebastian's answer help you? If so, please consider upvoting or accepting the answer. This is done by clicking the upwards normal distribution & possibly clicking on the check mark below the vote total. If it isn't quite what you needed, consider leaving a comment below his answer. $\endgroup$ – gung Oct 24 '18 at 14:55
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You can use that you already proved the result for $n <k+1$.

Therefore $$P(\cup_{i=1}^k(A_{k+1}\cap A_i))=\sum_{i=1}^k P(A_{k+1}\cap A_i)-\sum_{i<j}^k P((A_{k+1}\cap A_i)\cap(A_{k+1}\cap A_j))- \sum_{i<j<l}^kP((A_{k+1}\cap A_i)\cap (A_{k+1} \cap A_j) \cap (A_{k+1} \cap A_l))- ... - (-1)^k((A_{k+1}\cap A_1)\cap...\cap(A_{k+1} \cap A_k))$$

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