Assume that V = min(X, Y), where X and Y are independent random variables with the respective PDFs fX(x) = λe^-λx x ≥ 0 fY(y) = μe^-μy y ≥ 0 where λ 0 and μ 0. What is the PDF of V?

Question

Assume that [latex]V=\min(X, Y),[/latex] where [latex]X[/latex] and [latex]Y[/latex] are independent random variables with the respective PDFs

[latex]\,\displaystyle \begin{array}{ll}f_X\left (x ight )=\lambda e^{-\lambda x} & x \geq 0 \f_Y\left (y ight )=\mu e^{-\mu y} & y \geq 0\end{array}[/latex]

where [latex]\lambda \gt 0[/latex] and [latex]\mu \gt 0[/latex]. What is the PDF of [latex]V[/latex]?

Accepted Answer

We first obtain the CDFs of [latex]X[/latex] and [latex]Y[/latex] , which are as follows:

[latex]\,\displaystyle \begin{aligned}& F_X\left (x ight )=P[X \leq x]=\int_0^x \lambda e^{-\lambda u} d u=1-e^{\lambda x} \& F_Y\left (y ight )=P[Y \leq y]=\int_0^y \mu e^{-\mu w} d w=1-e^{-\mu y}\end{aligned}[/latex]

Thus, the PDF of [latex]V[/latex] is given by

[latex]\,\displaystyle f_V\left (v ight )=f_X\left (v ight )\left\{1-F_Y\left (v ight ) ight\}+f_Y\left (v ight )\left\{1-F_X\left (v ight ) ight\}\=\lambda e^{-\lambda v} e^{-\mu v}+\mu e^{-\mu v} e^{-\lambda v}=\lambda e^{-\left (\lambda+\mu ight ) v}+\mu e^{-\left (\lambda+\mu ight ) v}[/latex]

[latex]\,\displaystyle =\left (\lambda+\mu ight ) e^{-\left (\lambda+\mu ight ) v} \quad v \geq 0[/latex]

Since [latex]\lambda[/latex] and [latex]\mu[/latex] are the failure rates of the components, the result indicates that the composite system behaves like a single unit whose failure rate is the sum of the two failure rates. More importantly, [latex]V[/latex] is an exponentially distributed random variable whose expected value is [latex]E[V]=1/( \lambda + \mu )[/latex].

Question 6.14: Assume that V = min(X, Y), where X and Y are independent ran......

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