Distribution of the Median
by greenturtle3141, Apr 8, 2022, 8:21 PM
Reading Difficulty: 4/5
reals are chosen uniformly at random from the interval ![$[0,1]$](//latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. Let random variable 
be the median of these numbers. What is the distribution of ?
CLAIM: is a continuous random variable with density:
![$$f_n(x)=(2n+1)\binom{2n}{n}x^n(1-x)^n \cdot 1_{[0,1]}(x)$$](//latex.artofproblemsolving.com/a/c/2/ac2f8f057a32f76e49015b6f4d596596728325b5.png)
Proof. Let's try to compute the CDF,
. For 
we have that 
iff at least 
of the random reals chosen lie in ![$[0,t]$](//latex.artofproblemsolving.com/9/3/9/9398cf678fee12444e20e80c8ec38bb2803749fb.png)
. Such an event may be partitioned into the following events:
Computing the probability that exactly
of the numbers lie in may be reasoned as follows: There are 
ways to choose which of the numbers will lie in . Then the probability that these numbers lie in is 
, and the probability that the other 
numbers lie in ![$(t,1]$](//latex.artofproblemsolving.com/d/0/9/d09224ae5de063157ea0b178112e8b00db3d9ebb.png)
is 
. Altogether, this is a probability of 
.
It follows that:
Since this is (absolutely) continuous, we deduce that indeed is a continuous random variable and that a density exists.
You might think that we're now going to brutally expand this, differentiate, and then pray that the result is the claimed density... but some beautiful magic happens if we differentiate now:
Let's examine this first term:
Wait, isn't that the claimed density? Now we just need to show that everything else just zeroes out I guess... indeed, this series telescopes! To see why, just manipulate this expression, for 
:
Wow! So everything cancels except the very first term (the claimed density) and the very last term (which is just 
). 
For fun we can now compute the expectation and variance of.
CLAIM:
Proof. I mean, duh.
CLAIM:
Proof. A simple inductive argument shows that for all
, we have:
Applying this, we have:
And if you trust me, this ends up simplifying nicely to the claimed variance.
reals are chosen uniformly at random from the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. Let random variable 
be the median of these numbers. What is the distribution of ?CLAIM:
is a continuous random variable with density:![$$f_n(x)=(2n+1)\binom{2n}{n}x^n(1-x)^n \cdot 1_{[0,1]}(x)$$](http://latex.artofproblemsolving.com/a/c/2/ac2f8f057a32f76e49015b6f4d596596728325b5.png)
Proof. Let's try to compute the CDF,
. For 
we have that 
iff at least 
of the random reals chosen lie in ![$[0,t]$](http://latex.artofproblemsolving.com/9/3/9/9398cf678fee12444e20e80c8ec38bb2803749fb.png)
. Such an event may be partitioned into the following events:- Exactly of the numbers lie in .
- Exactly
of the numbers lie in . - ...
- Exactly of the numbers lie in .
of the numbers lie in Computing the probability that exactly
of the numbers lie in may be reasoned as follows: There are 
ways to choose which of the numbers will lie in . Then the probability that these numbers lie in is 
, and the probability that the other 
numbers lie in ![$(t,1]$](http://latex.artofproblemsolving.com/d/0/9/d09224ae5de063157ea0b178112e8b00db3d9ebb.png)
is 
. Altogether, this is a probability of 
.It follows that:
Since this is (absolutely) continuous, we deduce that indeed is a continuous random variable and that a density exists.You might think that we're now going to brutally expand this, differentiate, and then pray that the result is the claimed density... but some beautiful magic happens if we differentiate now:
Let's examine this first term:
Wait, isn't that the claimed density? Now we just need to show that everything else just zeroes out I guess... indeed, this series telescopes! To see why, just manipulate this expression, for 
:
Wow! So everything cancels except the very first term (the claimed density) and the very last term (which is just 
). 
For fun we can now compute the expectation and variance of
.CLAIM:
Proof. I mean, duh.
CLAIM:
Proof. A simple inductive argument shows that for all
, we have:
Applying this, we have:
And if you trust me, this ends up simplifying nicely to the claimed variance. 
March 2025
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