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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- Winter '08
- Probability theory, x3, X1, C.D. Cutler