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$$p(A|B) = \frac{p(A,B)}{p(B)}.$$ $A$ is the event that the die was fair. $B$ is the event that the $6$ die rolls turned out as specified. $$p(B) = p(A,B) + p(A^C,B).$$ $$p(A|B) = \frac{(1/2) \times (... 2 votes ### How to rigorously interpret and transform "equal chance" in different ways? Putting the 100 balls randomly in the 10 boxes is equivalent to choosing randomly a mapping from the set of balls to the set of boxes. There are 10^{100} such mappings. The event «no box is ... 2 votes Accepted ### Sequence of random variables converging to zero arbitrarily slowly For any \delta>0 there exists N_{\delta}\ge 1 such that$$ \mathsf{P}(X_n\ge \delta)\le \delta $$for all n\ge N_{\delta}. Thus, you may construct the required sequence \{a_n\} as follows. ... 1 vote ### Show that no two sets in the probability space with \mathbb{P}(\{k\})=2^{-k!} are independent. Non-existence of non-constant independent random variables on this space: Let X and Y be independent random variables on this space and suppose they are both non-constant. Let E be a non-empty ... 1 vote ### How to rigorously interpret and transform "equal chance" in different ways? First of all, re Stars and Bars Theory, the two methods that you used do give the same answer. That is p_{10}(100) bijects to determining the number of non-negative integer solutions to x_1 + \... 1 vote ### (X,\Sigma,\mu) be a probability space, A\subset\Sigma be a field s.t. \Sigma=\sigma(A).Prove inf\{\mu(E\Delta F):F\in A\}=0\forall E\in\Sigma Hints: Choose F_n \in \mathcal A such that \mu (E_n\Delta F_n)<\frac {\epsilon} {2^{n}}. Check that (\bigcup E_n) \Delta (\bigcup F_n )\subseteq \bigcup (E_n \Delta F_n). Of course, \bigcup ... 1 vote Accepted ### (X,\Sigma,\mu) be a probability space, A\subset\Sigma be a field s.t. \Sigma=\sigma(A).Prove inf\{\mu(E\Delta F):F\in A\}=0\forall E\in\Sigma I think one can prove directly that \mathcal{M} is a \sigma-algebra. Given \varepsilon>0, If A\in\mathcal{M} then there is A'\in\mathcal{A} such that \mu(A\triangle A')<\varepsilon. ... 1 vote Accepted ### Equivalence of a Levy measure For |x| \leq 1,$$|x|^2 - \frac{|x|^2}{1 + |x|^2} = \frac{|x|^4}{1 + |x|^2} \leq \frac{|x|^2}{1 + |x|^2} \in L^1(\nu).$$So the first integral is finite. For |x| \geq 1,$$1 - \frac{|x|^2}{1 + |x|^...
Your first application of Fubini's theorem is correct, but the second one does not seem to hold in general. First application Recall that to apply Fubini's theorem to $f$, you must have \$f \in L^1( \...