Two basic exercises on order statistics

[psie]

Homework Statement

Let $X_{(k)}$ denote the $k$th order variable of a sample of size $n$, where $X_1,\ldots,X_n$ is the sample from a distribution with distribution function $F$. Suppose that $F$ is continuous.

(a) Compute $P(X_k=X_{(k)},k=1,2,\ldots,n)$, that is, the probability that the original, unordered sample is in fact (already) ordered.
(b) Compute $P(X_{k;n}=X_{k;n+1})$, that is, the probability that the $k$th smallest observation still is the $k$th smallest observation. (Here $X_{k;n}$ is still the $k$th order variable, and the $n$ simply denotes the sample size.)

Relevant Equations

The number of permutations of $n$ distinct objects is $n!$.

Note, it's not assumed anywhere that $X_1,\ldots,X_n$ are independent.

My solution to (a) is simply $1/n!$, since we've got $n!$ possible orderings, and one of the orderings is the ordered one, so $1/n!$. However, I am not sure this is correct, since I don't understand why the assumption $F$ continuous is necessary. Grateful for an explanation.

For (b), my answer I think is not complete. I reasoned as follows; if we're given a sample of size $n$, the probability to find $X_k$ as $X_{k;n}$ is $1/n$, since there are again $n!$ orderings and in $(n-1)!$ of these we'll find $X_k$ as $X_{k;n}$, so $(n-1)!/n!=1/n$. This is all I got for (b). I think there's something missing.

 

[FactChecker]

However, I am not sure this is correct, since I don't understand why the assumption $F$ continuous is necessary.

If it was a discrete probability, you would have to worry about ties.

For (b), my answer I think is not complete. I reasoned as follows; if we're given a sample of size $n$, the probability to find $X_k$ as $X_{k;n}$ is $1/n$, since there are again $n!$ orderings and in $(n-1)!$ of these we'll find $X_k$ as $X_{k;n}$, so $(n-1)!/n!=1/n$. This is all I got for (b). I think there's something missing.

That seems logical to me.

 

[psie]

That seems logical to me.

Thanks! Part (b) seems kind of strange to me. Seems like a conditional probability, i.e. given that the sample of size $n$ is ordered, what is the probability that it still will be ordered when making one further observation? This is $1/n$.

EDIT: The probability that it will still be ordered is $1/(n+1)$. The statements I made concerning part (b) in post #1 are correct I think, but here I am making a different statement.

 

[psie]

I think there is another way of interpreting the question and I think this is the correct way. I don't know how else to put it other than the way the problem originally put it. What is the probability that the $k$th smallest observation is the $k$th smallest observation? This is the probability $P(X_{n+1}>X_{k;n})$. Consider $$\underbrace{\phantom{\quad}}_{\text{1st slot}}X_{1;n}\underbrace{\phantom{\quad}}_{\text{2nd slot}}X_{2;n}\quad\ldots\quad X_{n;n}\underbrace{ \ }_{(n+1)\text{st slot}}.$$Now, there are $(n-k+1)$ slots above $X_{k;n}$, so the probability that the $k$th smallest observation remains the $k$th smallest one is $\frac{n-k+1}{n+1}$.