21 this mapping maps r2 into the set which is

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Unformatted text preview: nnection with failure rate functions.) Solution: We compute FW first. A nice observation is that, for any constant t, max(X, Y ) ≤ t if and only if X ≤ t and Y ≤ t. Equivalently, the following equality holds for events: {max(X, Y ) ≤ t} = {X ≤ t} ∩ {Y ≤ t}. By the independence of X and Y, FW (t) = P {max(X, Y ) ≤ t} = P {X ≤ t}P {Y ≤ t} = FX (t)FY (t). Differentiating with respect to t yields fW (t) = fX (t)FY (t) + fY (t)FX (t). (4.20) There is a nice interpretation of (4.20). If h is a small positive number, the probability that W is in the interval (t, t + h] is fW (t)h + o(h), where o(h)/h → 0 as h → 0. There are three mutually exclusive ways that W can be (t, t + h]: Y ≤ t and X ∈ (t, t + h] : has probability FY (t)fX (t)h + o(h) X ≤ t and Y ∈ (t, t + h] : has probability FX (t)fY (t)h + o(h) X ∈ (t, t + h] and Y ∈ (t, t + h] : has probability fX (t)hfY (t)h + o(h) So fW (t)h = FY (t)fX (t)h + FX (t)fY (t)h + fX (t)fY (t)h2 + o(h). Dividing this by h and letting h → 0 yields (4.20). 4.7 Joint pdfs of functions of random...
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