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MixedContDiscreteExample

# MixedContDiscreteExample - use them rather than a...

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C.D. Cutler STAT 333 Mixed Discrete-Continuous Conditioning Example Suppose X 1 , X 2 , X 3 are i.i.d. exponential random variables with rate parameter λ > 0. Find P ( A ) where A = [ X 1 less than both X 2 , X 3 ]. method one by symmetry: Since X 1 , X 2 , X 3 are i.i.d. (hence no variable is favoured over another) each of the possible 3!=6 orderings [ X j 1 < X j 2 < X j 3 ] is equally likely to occur. Hence each ordering has probability 1/6 of occurring. Thus P ( A ) = P ([( X 1 < X 2 < X 3 ]) + P ([( X 1 < X 3 < X 2 ]) = 1 / 6 + 1 / 6 = 1 / 3 . method two by an even easier symmetry method: Since X 1 , X 2 , X 3 are i.i.d. any one of them is equally likely to be the smallest of the lot. Hence, directly, P ( A ) = 1 / 3. Note that neither of the above symmetry methods used the fact that the variables were exponentially-distributed, only that they were i.i.d. The above result holds for any continu- ous i.i.d. random variables. The symmetry proofs are so simple here that of course we would use them rather than a conditioning argument.
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Unformatted text preview: use them rather than a conditioning argument. However, we carry out the conditioning argument for the exponential case only because the approach here sets us up for a proof of the Alarm Clock Lemma much later in the course (where the variables are independent but possibly non-identically distributed exponentials). method three by conditioning on X 1 = x : P ( A ) = E ( I A ) = i ∞ P ([ X 3 > X 1 , X 2 > X 1 ] /X 1 = x ) f 1 ( x ) dx = i ∞ P ([ X 3 > x, X 2 > x ] /X 1 = x ) f 1 ( x ) dx by substitution = i ∞ P ([ X 3 > x, X 2 > x ]) f 1 ( x ) dx using X 3 , X 2 independent of X 1 = i ∞ P ([ X 3 > x ]) P ([ X 2 > x ]) f 1 ( x ) dx using X 3 and X 2 independent = i ∞ e-λx e-λx λe-λx dx using exponential tails and p.d.f. = i ∞ λe-3 λx dx = 1 3 (0.1) 1...
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