Exponential Distribution, Statistics. I know X has an exponential distribution with parameter $\theta =2$. 
I was asked to define $Y=lnx$ and determine the suppose of Y and the pdf for Y. Then let $X_1, X_2$ be two independent observations from X and find P(max X_i <1)
I have no idea where to begin. Any help would be awesome!
Thanks all!
 A: The meaning of exponential distribution with parameter $\theta$ varies. Some would say that the density function is $\frac{1}{\theta}e^{-x/\theta}$ (for $\theta\gt 0$). Some would say that the density function is $\theta e^{-x\theta}$. 
If the parameter is called the mean, then it is  $\frac{1}{\theta}e^{-x/\theta}$. You will have to check what convention your book/course uses. To bypass this awkwardness, I will assume that the density function is $\lambda e^{-\lambda x}$.  Depending on what convention the book/course uses, we have $\lambda=\frac{1}{2}$ or $\lambda=2$. On the basis of an earlier question that I saw that seemed to come from the same exercise set, I suspect that $\lambda=\frac{1}{2}$.   After these lengthy preliminaries, on to the questions!

Look at the random variable $Y$, where $Y=\ln X$. As $X$ travels over the interval $(0,\infty)$, $Y$ travels over the interval $(-\infty, \infty)$.
We find the cumulative distribution function (cdf) $F_Y(y)$ of $Y$. We have
$$F_Y(y)=\Pr(Y\le y)=\Pr(\ln X\le y)=\Pr(X\le e^y)=\int_0^{e^y}\lambda e^{-\lambda x}\,dx=1-e^{-\lambda e^y}.$$
For the density function $f_Y(y)$, differentiate. We get $\lambda e^y e^{-\lambda e^y}$.
For the probability that $\max(X_1,X_2)\lt 1$, we need to find the probability that both $X_1$ and $X_2$ are $\lt 1$. By independence, this is $\Pr(X_1\lt 1)\Pr(X_2\lt 1)$. Finally, note that for example $\Pr(X_1\lt 1)=\int_0^1 \lambda e^{-\lambda x}\,dx=1-e^{-\lambda}$.   
