Let $X_1,\cdots,X_n$ be independent and identically distributed random variables with distribution $F$. We say that a record occurs at time $n,n>0$ and has value $X_n$ if $X_n > \max (X_1,\cdots X_{n-1})$
Suppose $X_n$ is a record value. Will $F(X_n)$ be a record value as well? My book says that the record times of the sequence $\{X_1,\cdots,X_n\}$ will be the same as for the sequence $F(X_1),\cdots,F(X_n)$
Attempt: $X_n$ is a record value implies $X_n > \max (X_1,\cdots X_{n-1})$. Let the value obtained at time $t$ be $x_n$
We need to find if $F(X_n) > \max (F(X_1),\cdots,F(X_{n-1}))$
Now, $F(X_n = y) = \mathbb P(X_n \leq y) = \int_{-\infty} ^{y} f_X(x) dx$
$(X_i)'s$ are i.i.d random variables.
$F(X_n)$ will be a record value if $F(X_n) > \max (F(X_1),\cdots,F(X_{n-1}))$ . We are given that $(X_n) > \max ((X_1),\cdots,(X_{n-1}))$
At time $n, F(X_n) \equiv F(X_n = x_n) = \int_{-\infty} ^{x_n} f_X(x) dx$
Since $x_n > \max (x_1,\cdots,x_{n-1}) \implies \int_{-\infty} ^{x_n} f_X(x) dx \ge \int_{-\infty} ^{x_i} f_X(x) dx$
$ \implies F(X_n)$ must be the record value as well for $i<n$.
Is this correct?