You are given two identical components. We note X1 the lifetime of component 1,

Question

You are given two identical components. We note X1 the lifetime of component 1, and X2 the lifetime of component 2. X1 and X2 are independent exponential random variables, of parameter 1/2. we recall the p.d.f. of an exponential R.V. of parameter lambda: f(x) = lambda e- lambda x, for x 0) Write the p.d.f. of X1, of X2, and the joint p.d.f. of X1,X2. Compute P(X1 > 2X2). We note T1 the first time when one component breaks (T1 = min(X1, X2)). Compute the c.d.f. of T1, and deduce its p.d.f. T2 is the time when the last surviving component breaks (T2 = max(X1,X2)). Compute the c.d.f. of T1, and deduce its p.d.f. Compute joint distributions (c.d.f. and p.d.f.) of T1 and Ti. Compute P(T2 > 2T1).

Explanation / Answer

This is the complete solution

pdf1=lambda1e^(-lambda1*x)

pdf2=lambda2e^(-lambda2*x)

as both distribution as independent of each other hence

pdf joint func=lambda1e^(-lambda1*x)*lambda2e^(-lambda2*x)

F(x)=P(X<x)=T=min(X1,X2)<x(say)

=1-P((x1>x)intersectn(x2>x))

=1-P(x1>x)*P(x2>x) (independent)

=1-[1-P(x1<x)][1-P(x2<x)]

=2F(x)-F(x)^2 (cpdf)

f(x)=d(F(x))dx=2f(x)-2f(x)F(x)

do similarly for fourth part

5) for fifth part just write the pdfs of T1 and T2 and multiply them to get the joint distribution.

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