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$X_1$, $X_2$,...,$X_n$ are $n$ i.i.d. normal random variables. $Y$ is the median of these variables. I am asking if for a finite even number $n$, $X_1-Y$, $X_2-Y$, ..., $X_n-Y$ are i.i.d. symmetric distributed random variables, so that the probability of $X_i-Y>0$ equals $0.5$.

Let $n=100$, my simulation results (with $10^4$ repetition) show that there is almost no difference of the skewness between $X_i$ and $X_i-Y$ for $i=1,...,n$. Both skewnesses center around zero.

Kefu Liao
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  • Did you try to answer that yourself analitically? – Aig Mar 21 '24 at 17:50
  • From Wikipedia, I know that median $Y$ is approximately normal. If $X_i$ is independent of $Y$, $X_i-Y$ should be normal thus a symmetric distribution. But I have no idea if $X_i$ and $Y$ are independent. – Kefu Liao Mar 21 '24 at 18:05
  • If you draw a finite number of $X_i$ from a probability distribution they are just numbers and no longer random variables. It is not surprising that their histogram can be made symmetric if it was not already. – Kurt G. Mar 21 '24 at 18:06
  • I do simulations with $10^4$ repetitions for proxying the distribution of these random variables. – Kefu Liao Mar 21 '24 at 18:10
  • You are asking if for a finite $n,,$ $X_1-Y,,\dots, X_n-Y$ are i.i.d symmetric distributed random variables. I said that they are not random variables and that it is not surprising that $Y$ is nearly zero and that they are symmetric. – Kurt G. Mar 21 '24 at 18:15

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By exchangeability of $X_1,X_2,\ldots,X_n$, the $X_1-Y,X_2-Y,\ldots, X_n-Y$ are identically distributed to each other.

Comparing the joint distribution of $(X_1-Y,X_2-Y,\ldots, X_n-Y)$ with that of $(-X_1+Y,-X_2+Y,\ldots, -X_n+Y)$ and using the symmetry of normal distributions, they are each symmetrically distributed about $0$.

They are not independent: for example if $n$ is odd then (with probability $1$) exactly one of $X_1-Y,X_2-Y,\ldots, X_n-Y$ is zero so each has probability $\frac1n$ of being $0$, and exactly half of the others are negative and the rest positive; with even $n$ and taking the median as halfway between the middle two values, exactly half are negative and half positive. So knowing the value of one $X_i-Y$ gives you some information about the distribution of the others.

This also means that with odd $n$ you have $\mathbb P(X_i -Y>0)=\frac{\frac{n-1}2}{n}=\frac12-\frac1{2n}$, while with even $n$ and taking the median as halfway between the middle two values you have $\mathbb P(X_i -Y>0)=\frac{\frac{n}2}{n}=\frac12$.

Henry
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  • Why each of $X_1,X_2,...,X_n$ is symmetrically distributed about $0$? Could you give a proof using the definition of the symmetrical distribution? – Kefu Liao Mar 21 '24 at 19:55
  • @KefuLiao Each of $X_1, \ldots, X_n$ is symmetrically distributed about the mean of its normal distribution and $X_1-\mu$ has the same distribution as $\mu-X_1$. Similarly each of $X_1-Y, \ldots, X_n-Y$ is symmetrically distributed about $0$ as $X_1-Y$ has the same distribution as $-X_1+Y$. Note that if the median of $X_1, \ldots, X_n$ is $Y$ then the median of $-X_1,\ldots,-X_n$ is $-Y$. – Henry Mar 21 '24 at 22:44