# 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!

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 $\Pr(X_1\lt 1)=\int_0^1 \lambda e^{-\lambda x}\,dx=1-e^{-\lambda}$.

• Thanks! I thought from this I would be able to figure out the rest of the problem set, but it seems like I cant. If we have x_1, x_2...x_36 independent observations, how would I find the mean and variance of $\bar{X} = \sum_{i=1}^{36} \frac{1}{36}X_i$ Commented May 12, 2014 at 14:15
• You are welcome. It is best to do it with general facts that are undoubtedly done in your course. The mean of a sum is the sum of the means, the mean of $kW$ is k$times the mean of$W$. Putting things together, we get that the mean of$\bar{X}$is$\frac{1}{36}$times the sum of the means of the$X_i$. If your$X_i$have mean$2$, that makes the mean of$\bar{X}$also equal to$2$. This if you think about it makes intuitive sense. (Cont. in anoter comment for variance) Commented May 12, 2014 at 15:08 • (Cont) The above is one reason one should not answer in comments! Summary: The mean of$(1/36)\sum X_i$is$(1/36)$times the sum of the means. In our case that is$2$. For the variance you probably know that the variance of the$X_i$is$2^2$. The variance of an independent sum is the sum of the variances, and the variance of$kW$is$k^2$times the variance of$W$. This gives that the variance of$\bar{X}$is$(1/36)^2(36)(2^2)=\frac{2^2}{36}\$. Commented May 12, 2014 at 15:14
• To anyone who can delete (or even better, fix) the comment that starts with "You are welcome," or tell me how to do it, help! Commented May 12, 2014 at 15:20