Do binomial distributions $Bin(n,p)$ always have increasing hazard rate?
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If the sample size reduces with each failure, does the hazard rate, by definition, increase? (It seems obvious, so maybe I am missing something here - can you clarify the question?) – Red Five Apr 24 '24 at 11:15
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The hazard rate is defined as $\mathbb{P}(X=k)/\mathbb{P}(X\geq k)$. For $X$ binomial I find it very difficult to show that it is an increasing function, for all $n$ and $p$, so maybe it is not? – xyz Apr 24 '24 at 11:20
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@RedFive My intuition is exactly as you say, but it is informal. I think it should be shown analytically though, if it is true. – xyz Apr 24 '24 at 11:27
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Empirically it seems true, increasing from $(1-p)^n$ (when $k=0$) to $1$ (when $k=n$). For $k < np$, the numerator is increasing while the denominator is decreasing. – Henry Apr 24 '24 at 13:35
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Again empirically, if $h(k) = \frac{\mathbb{P}(X=k)}{\mathbb{P}(X\geq k)}$, it seems you can say $h(k) \ge \left(1+\frac{p}{n(1-p)}\right) h(k-1)$ for $1 \le k \le n$ and $0 \lt p \lt 1$, which would imply $h(k)$ is increasing. – Henry Apr 24 '24 at 14:05