Imagine I have a real random variable $X$ with some distribution (continuous, discrete or continuous with atoms)

Now Imagine I have i.i.d. copies $X_1,...,X_n$, all independently and equally distributed as $X$

My claim is:

$$\mathbb{P}(X_2>X_1)=\mathbb{P}(X_2<X_1)$$ My secondy claim is the following:If I order them by size, so that $X_{(1)}<X_{(2)}<\ldots<X_{(n)}$ and I define the interval $I_n=[X_{(1)},X_{(n)}]$; Then I claim:


So the probability that the $n+1$-th number exceeds the interval on the left equals the probability it exceeds on the right

I guess the first one is true, but the second one not;

E.g. Assume X can take the value 0 and 1; and assume $n=3$; Then

$$\mathbb{P}(X_3>{X_1,X_2})=\mathbb P (X_3=1)\mathbb P (X_2=0)\mathbb P (X_1=0)=\mathbb P (X=1)\mathbb P (X=0)\mathbb P (X=0)$$ but also $$\mathbb{P}(X_3<{X_1,X_2})=\mathbb P (X_3=0)\mathbb P (X_2=1)\mathbb P (X_1=1)=\mathbb P (X=0)\mathbb P (X=1)\mathbb P (X=1)$$

which is gernerally not the same; But What I am wondering is if there are simply conditions that it would become true

  • $\begingroup$ Your notation looks rather confusing to me. First "I generate copies of $X$"... I guess you mean different realizations of the random variable . Or more standard: $X_i$ are iid random variables. Afterwards you denote by the same $X_i$ the ordered (ranked) variables - this is confusing, and more so when you write the event $(X_{n+1}<X_1)$ . $\endgroup$ – leonbloy Sep 11 '18 at 13:18
  • $\begingroup$ I'll correct it, thank you $\endgroup$ – user299124 Sep 11 '18 at 14:35

Let $Y_n =\min(X_1,X_2 \cdots X_n)$, and let $y_n=\sum_{i=1}^n[X_i=Y_n]$ count the number of elements that attain that minimum. Analogously, let $Z_n$ and $z_n$ be the maximum and maximum-count.

Then, by symmetry $P( X_{n} = Y_n \wedge y_n=1)=P(X_n=Y_n) P(y_n=1 \mid X_n=Y_n)=\frac{1}{n} P(y_n=1)$

Then, esentially you are asking if $P(y_n=1)=P(z_n=1)$ , that is, if the probability of having a single maximum equals the probability of having a single minimum. This is not true in general.

It's true for a continuous variable (continuous CDF) because in that case the probability of a having a single extrema equals $1$. It's also true for a symmetric (around the median) random variable. I'm not sure if there's a simple characterization for its CDF to be true in general.


Let $F(x) = P(X \le x)$ be the CDF, and let $p(x)= F(x) - F(x^-)$.

Then the probability of having a single minimun in $n+1$ realizations equals

$$A=p(y_{n+1}=1)= \int \left(\frac{1-F(x)}{1-F(x^-)}\right)^n dF(x)= \int \left(1-\frac{p(x)}{1+p(x)-F(x)}\right)^n dF(x) \tag{1}$$

Similarly, for the maximum:

$$B=p(z_{n+1}=1)= \int \left(\frac{F(x^-)}{F(x)}\right)^n dF(x) = \int \left(1- \frac{p(x)}{F(x)}\right)^n dF(x) \tag{2}$$

If $F(x)$ have finite discontinuities at $x_i$, $i=1,2\cdots k$ (perhaps the result is also valid for more general settings), we can write $F(x)=F_c(x) + \sum_i p(x_i)u(x-x_i)$ where $F_c(x)$ is continuous and $u(\cdot)$ is the unit-step function. Then

$$\begin{align} A &=\sum_i p(x_i) \left(1-\frac{p(x_i)}{1+p(x_i)-F(x_i)}\right)^n +F_c(+\infty)\\ &=1- \sum_i p(x_i)\left[1- \left(1-\frac{p(x_i)}{1+p(x_i)-F(x_i)}\right)^n \right]\tag{3} \end{align} $$

$$\begin{align} B&=\sum_i p(x_i) \left(1- \frac{p(x_i)}{F(x_i)}\right)^n +F_c(+\infty)\\ &=1- \sum_i p(x_i)\left[1- \left(1- \frac{p(x_i)}{F(x_i)}\right)^n \right] \tag{4} \end{align}$$

Of course, $A=B=1$ if $F(x)$ is continuous. Also, $A=B$ if the probability (both the continuous and the discrete parts!) is symmetric. There's not much more to say in general, I think...

  • $\begingroup$ Thank you, that was very smart - now I only need to figure out if it is true if there is an atom somewhere in the support of $X$ but not at the beginning or end of the support... $\endgroup$ – user299124 Sep 11 '18 at 14:34

You're asking whether $P(X > \max({X_1, \ldots, X_n})) = P(X < \min({X_1, \ldots, X_n}))$, where $X_1, \ldots, X_n$ and $X$ are all independent and distributed the same.

This is true in some distributions and false in others. For example if the $X_i$s are sampled uniformly at random from $[0,1]$ then the probabilities would be the same (due to symmetry).

In general, for any distribution over $[a,b]$ which is symmetric about $(a+b)/2$, you would expect this to be true.

Also, I believe this is true for any continuous distribution over $(a,b)$ (i.e. such that the probability of any two variables receiving exactly the same value is 0). To see this, note that $P(X > \max({X_1, \ldots, X_n})) = \frac{1}{n+1} = P(X < \min({X_1, \ldots, X_n}))$. This is because we can first generate $X, X_1, \ldots, X_n$, then order $X_1, \ldots, X_n$, without affecting the probabilities, and any of these are equally likely to be the maximum or the minimum.

To see when this could be false, suppose that the $X_i$s are chosen as follows: with probability $1/2$, set $X_i = 10$, otherwise sample $X_i$ uniformly from $[0,1]$. Then $P(X > \max({X_1, \ldots, X_n}))$ will approach $0$ exponentially fast, since it is impossible for $P(X > 10)$ to happen, but $P(X < \min({X_1, \ldots, X_n}))$ is roughly proportional to $1/n$.

  • $\begingroup$ Thank you for your answer... Well I do, the 2nd statement is equivalent to: $\mathbb{P}(x_{n+1}<\min(x_1,...,x_n))$ and $\mathbb{P}(x_{n+1}>\max(x_1,...,x_n))$; The statement does I think not hold but it feels like the statement could hold assuming $X$ to follow some distributional properties $\endgroup$ – user299124 Sep 11 '18 at 11:29
  • $\begingroup$ @J.Doe Sorry, I misread your question. Did I understand you correctly this time? $\endgroup$ – user97678 Sep 11 '18 at 11:42
  • $\begingroup$ You did - I had lots of flaws writing it in a real way, sorry for that; But the thing is: When does it hold, e.g. does it hold when the support of $X$ does not start or end with an atom?Like in your or my example, the highest or lowest number was an atom - what if it were not? Or what if the distribution is just continuous? $\endgroup$ – user299124 Sep 11 '18 at 11:45
  • $\begingroup$ @J.Doe Seems we had the same thoughts - I don't know of anything more general than this. For "atoms in the middle" I believe you can do the same trick in my answer (with the "10"), by simply making it extremely unlikely for any variable to be larger than the atoms. $\endgroup$ – user97678 Sep 11 '18 at 12:22
  • $\begingroup$ @J.Doe By the way, note that your derivation in the comment above is not quite correct: $X > X_i$ and $X > X_j$ are not independent! $\endgroup$ – user97678 Sep 11 '18 at 12:40

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