We say that ζ is a p-percentile of the distribution F if F(ζ) = p. Show that if ζ is a p-percentile of the IFRA distribution F, then
$\bar{F} (x)= ≤ e^{−θx}, x ≥ζ\\ \bar{F} (x) ≥ e^{−θx}, x ≤ζ$
where
$θ = \frac{ −log(1 −p)}{ζ}$

Question

We say that ζ is a p-percentile of the distribution F if F(ζ) = p. Show that if ζ is a p-percentile of the IFRA distribution F, then

[latex]\bar{F} (x)= ≤ e^{−θx}, x ≥ζ\ \bar{F} (x) ≥ e^{−θx}, x ≤ζ[/latex]

where

[latex]θ = \frac{ −log(1 −p)}{ζ}[/latex]

Accepted Answer

For [latex]x \geqslant \xi[/latex],
[latex]1-p=\bar{F}(\xi)=\bar{F}(x(\xi / x)) \geqslant[\bar{F}(x)]^{\xi / x}[/latex]
since IFRA. Hence, [latex]\bar{F}(x) \leqslant(1-p)^{x / \xi}=e^{- heta x}[/latex].
For [latex]x \leqslant \xi[/latex],
[latex]\bar{F}(x)=\bar{F}(\xi(x / \xi)) \geqslant[\bar{F}(\xi)]^{x / \xi}[/latex]
since IFRA. Hence, [latex]\bar{F}(x) \geqslant(1-p)^{x / \xi}=e^{- heta x}[/latex].

Question 9.EX.25: We say that ζ is a p-percentile of the distribution F if F(ζ......

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