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Let $X_1, \ldots , X_n$ be a set of independent and identically distributed continuous random variables. Show that the random vector $\bigl(X_n, X_{(n)}\bigr)^T$does not have a joint pdf.

I know for a vector $\bigl(X_n, X_{(n)}\bigr)^T$ to not have a joint pdf, it must be singular (ie, it's covariance matrix must not be of full rank). However, then I need to show this vector is singular. I know to be singular, $p > n$, which is $p=2$ in our case. How can I further prove that $\bigl(X_n, X_{(n)}\bigr)^T$ is singular (in order to prove it does not have a joint pdf)?

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It is not true that singular distributions never have a covariance matrix of full rank. For example, consider the pair $(X_1,X_2)$ where $X_1, X_2 \sim \operatorname{i.i.d. Bernoulli}(1/2).$ The distribution is discrete and therefore singular, but the covariance matrix is not singular.

By symmetry $$ \Pr(X_{(n)} = X_1) = \Pr(X_{(n)} = X_2) = \Pr(X_{(n)} = X_3) = \cdots = \Pr(X_{(n)} = X_n) $$ and thus each has probability $1/n.$

In the $(x,y)$-plane, the measure of the set $\{(x,y):x=y\}$ is $0;$ therefore the integral of every density function over that set is $0.$ But the probability that the pair $(X_{(n)}, X_n)$ is in that set is $1/n.$

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