Difference between revisions of "Notes:Distribution of the sample median"
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Revision as of 07:30, 11 December 2017
[ilmath]\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }[/ilmath]
[ilmath]\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } } [/ilmath][ilmath]\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } } [/ilmath][ilmath]\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} } [/ilmath]
Contents
Problem overview
Let [ilmath]X_1,\ldots,X_{2m+1} [/ilmath] be a sample from a population [ilmath]X[/ilmath], meaning that the [ilmath]X_i[/ilmath] are i.i.d random variables, for some [ilmath]m\in\mathbb{N}_{0} [/ilmath]. We wish to find:
- [math]\P{\text{Median}(X_1,\ldots,X_{2m+1})\le r} [/math] - the Template:Cdf of the median.
Initial work
Since the variables are independent then any ordering is as likely as any other (which I proved the long way, rather than just jumping to [math]\frac{1}{(2m+1)!} [/math] - silly me) however the result, found in Probability of i.i.d random variables being in an order and not greater than something will be useful.
