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say I have $X \sim \texttt{Binomial}(n, p)$ and $Y \sim \texttt{Binomial}(cn, p)$ where $0 \leq c \leq 1$. Also, I know what $E[W_{X_1}], E[W_{X_2}, ...]$ where $W_{X_i}$ is the $i$th largest sample. I was wondering if there was a straight forward way to get $E[W_{Y_i}]$ where $W_{Y_i}$ is the $i$th largest sample of $Y$?

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  • $\begingroup$ I assume $\frac{1}{c}$ divides $n$, otherwise $cn$ wouldn't be a whole number, right? $\endgroup$
    – Daniel P
    Mar 8, 2022 at 12:55
  • $\begingroup$ @DanielP yep, exactly. It may be more like round($cn$) but for now, I think it is fine to assume that $cn$ is a whole number. $\endgroup$
    – allyouneed
    Mar 8, 2022 at 13:45
  • $\begingroup$ I doubt it is straightforward. $\endgroup$
    – Henry
    Mar 8, 2022 at 16:45
  • $\begingroup$ As an example, suppose your sample sizes were $4$ and $X\sim \text{Bin}(4,\frac12)$ and $Y \sim \text{Bin}(2,\frac12)$, I see no obvious way to get from $E[X_{(1)}]\approx 1.01$, $E[X_{(2)}]\approx 1.70$, $E[X_{(3)}]\approx 2.30$, $E[X_{(4)}]\approx 2.99$ to $E[Y_{(1)}]\approx 0.32$, $E[Y_{(2)}]\approx 0.79$, $E[Y_{(3)}]\approx 1.21$, $E[Y_{(4)}]\approx 1.69$ $\endgroup$
    – Henry
    Mar 9, 2022 at 0:41
  • $\begingroup$ Yeah @Henry, I was struggling here as well. I could also do with a loose-ish lower bound though. If I had a lower bound on the expectation, that would work. For now I can get a lower bound of something like $E[W_{Y_i}] \geq E[W_{X_i}] - (1- c) \cdot n$. But, this is not good enough and seems to be too strong. $\endgroup$
    – allyouneed
    Mar 9, 2022 at 14:47

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