# Probaility of one random value being bigger than the other?

#### Maverick85

##### New Member
Hi all!
Intuition was telling me that probability of one random value being bigger than the other (both values are sampled from same distribution space) is 0.5. I did some simple tests of this with normally and uniformly distributed values and the results support my expectations.
Now, what I am not sure is if this is 'universal' truth? Does it hold true for all possible distributions? Any difference whether we are dealing with continuous, discrete or empirical distribution? Is there any paper or book that would explicitly show this - I need a better reference than 'my intuition'

#### vinux

##### Dark Knight
Is the "bigger than" is "Greater than" or "Greater than or equal to" ?

Anyway in continous case it doesn't matter . And your finding is true.. if the random variables are independent. And proof is simple..
( P(X<Y ) + P(X>Y) + P(X=Y) = 1
=> P(X<Y) + P(X>Y) =1 => 2*P(X<Y) =1 {since X & Y are iid } hence the result )

It may not be true,when it is discrete.. you can take any example are try this

#### Maverick85

##### New Member
Sorry, I have just not noticed that I made a crucial mistake in formulating the question:
Instead of one value being bigger than the other, I should write 1st being greater than the 2nd or with notation: P(X1>X2) . So, this is more restrictive. I hope this explains clearly what I am interested in...

@ vinux: but it would be true for some of discrete distributions as well right? Like uniform discrete...
"if the random variables are independent" - since both values are sampled from the same distribution space, they are independent, right?

Last edited:

#### BGM

##### TS Contributor
Usually we assume that the random sample are i.i.d., so X1 and X2 are i.i.d. too.
For discrete case, as vinux explained above,
we have Pr{X1 = X2} > 0
so Pr{X1 > X2} = Pr{X1 < X2} = (1 - Pr{X1 = X2})/2 < 1/2

#### Maverick85

##### New Member
I would like to build upon what we already discussed. Namely, I have a case where random values are obtained by permutation: I have 10 different values, I permute them and only look at first two values. Since in case of permutation I can not get the same values for both first and second values, I guess the logic of continuous distributions apply as well (when I do many permutations)? So, in each run P(X1=X2)=0 and over many runs expected probability P(X1>X2)=0.5.
Would you agree with my reasoning above or not? This is really important for me, so I would be extremely thankful for further discussion...

#### BGM

##### TS Contributor
OK then in this case X1 X2 are not identically distributed anymore
And you can simply argue that Pr{X1 < X2} = Pr{X1 > X2} = 1/2 by symmetry

Or more explicitly, let {x1, x2, ..., xn} be n distinct support points for X1
where x1 < x2 < ... < xn
Then Pr{X1 = xi} = 1/n, i = 1, 2, ..., n
Pr{X2 = xj|X1 = xi} = 1/(n - 1), j = 1, 2, ..., n, j ≠ i
⇒ Pr{X2 < X1|X1 = xi} = (i - 1)/(n - 1)

Thus Pr{X2 < X1}
= ∑Pr{X2 < X1|X1 = xi}Pr{X1 = xi}
= ∑Pr{X2 < xi}Pr{X1 = xi}
= ∑Pr{X2 < xi}/n
= ∑(i - 1)/[n(n-1)]
= [n(n+1)/2 - n]/[n(n-1)]
= (n+1-2)/[2(n-1)]
= 1/2

And by definition, Pr{X2 = X1} = 0 for i = 1, 2, ..., n
So Pr{X2 > X1} = Pr{X2 < X1} = 1/2
(you can also try to evaluate the series)