Distribution of Records Consider $Y_1, Y_2, \ldots Y_n, \ldots $ are i.i.d. continuous random variables with some CDF F. Define $Y_i$ to be a record if $Y_i > Y_j$ for all $j < i$. Let $R_1, R_2, \ldots R_i, \ldots $ denote the $i$th record among the $Y_i$'s. Find the CDF of $R_n$, as a function of $F$. 
I'm a bit stuck on how to start here. I tried calculating $P(R_n > x)$ and conditioning on whether $R_{n-1} > x$, but got stuck with this approach. 
 A: First consider this problem for exponentially distributed random variables. If $Y'_1,Y'_2,\ldots $ are i.i.d. exponentially distributed with mean $1$, it is not difficult to show that $R'_n$ has Gamma distribution with pdf 
$$
f_{R_n'}(x) = \dfrac{1}{(n-1)!}x^{n-1} e^{-x} \mathbb 1_{x>0}.
$$
After that random variables $Y_1,Y_2,\ldots$ which are i.i.d. continuous with CDF $F$ can be transformed to exponential $Y_i'$ by 
$$
Y_i'=-\ln(1-F(Y_i)).
$$
Note that this tranformation is monotone and it preserve inequalities between random variables, so any record value for initial sequence tranformed to the record value for the sequence of exponential r.v.'s. 
And 
$$
\begin{split}
\mathbb P(R_n \leq x) & = \mathbb P\bigl(-\ln(1-F(R_n))\leq -\ln(1-F(x))\bigr) \cr
& = \mathbb P\bigl(R_n'\leq -\ln(1-F(x))\bigr) = \int\limits_0^{-\ln(1-F(x))} \dfrac{1}{(n-1)!}x^{n-1} e^{-x}\, dx.
\end{split}
$$
Or the r.h.s. can be written as $\Gamma_{n,1}(-\ln(1-F(x)))$ where $\Gamma_{n,1}$ is CDF of Gamma distribution with shape $n$ and rate $1$. 
I am sure that there exist simple ways to show that records of a sequence of i.i.d exponential random variables have Gamma distribution. But they don’t occur to me, so I will use induction. I leave you the basis of induction. Let us make induction step. 
Assume that $R_{n-1}$ has pdf 
$$
f_{R_{n-1}}(y) = \frac{1}{(n-2)!}y^{n-1} e^{-y}\mathbb 1_{y>0}.
$$
Then find pdf of $R_n$. For $0<y<x$,
$$
f_{R_n,R_{n-1}}(x,y) dxdy = \mathbb P(R_n\in (x,x+dx), R_{n-1}\in (y,y+dy))
$$
Let $L_{n-1}$ is the moment of $n-1$st record :
$$
\{L_{n-1}=i\} = \{R_n=Y_i\}.
$$
Note that the distribution of the sequence $Y_{L_{n-1}+1},Y_{L_{n-1}+2},\ldots$ is the same as the distribution of the initial sequence. I will also write $R_n=x$ instead of $R_n\in(x,x+dx)$ for shortness.
$$
f_{R_n,R_{n-1}}(x,y) dxdy =\mathbb P(R_{n-1} =y, Y_{L_{n-1}+1}=x) +\mathbb P(R_{n-1} =y, Y_{L_{n-1}+1}<y, Y_{L_{n-1}+2}=x)+\ldots
$$
$$
=\mathbb P(R_{n-1} =y, Y_1=x) +\mathbb P(R_{n-1} =y, Y_1<y, Y_2=x)+\ldots
$$
$$
=f_{R_{n-1}}(y) f_{Y_1}(x) \sum_{k=0}^\infty F_{Y_1}^k(y)dxdy = f_{R_{n-1}}(y) f_{Y_1}(x) \frac{1}{1-F_{Y_1}(y)}\, dx\,dy
$$
$$
=\frac{1}{(n-2)!}y^{n-1} e^{-y} e^{-x} \cdot e^y\, dx\,dy=\frac{1}{(n-2)!}y^{n-1} e^{-x} \, dx\,dy.
$$
Then 
$$
f_{R_n}(x) = \int_0^x f_{R_n,R_{n-1}}(x,y)\, dy =\int_0^x \frac{1}{(n-2)!}y^{n-1} e^{-x} dy = \frac{1}{(n-1)!}x^{n-1} e^{-x}.
$$
