# Math Help - Confused about rand. var. transforms

1. ## Confused about rand. var. transforms

I have a problem where the rand var X= -(1/theta) ln(U) is given where U is uniform rand.var on (0,1) I need to prove that X has an exponential distribution function with parameter theta. I know i have to set MGF's equal to each other? but im not sure which formulas to work with. I notice that if I replace the X in the density function for an exponential RV with the X=-(1/theta) ln(U) it reduces to theta*(U). Not sure how this helps but it seemed like too much of a coincidence. I thought if i took the expected value of an exponential(1/theta) it would equal X= -(1/theta) ln(E(U)) but E(U) is 1/2 and therefore X=(1/theta)* ln(0.5). plz help me understand how to properly approach this, i have been working on it for hours, but my teachers notes dont cover it and i cant find a reliable help online

2. Hello,
Originally Posted by nine4fours
I have a problem where the rand var X= -(1/theta) ln(U) is given where U is uniform rand.var on (0,1) I need to prove that X has an exponential distribution function with parameter theta. I know i have to set MGF's equal to each other? but im not sure which formulas to work with. I notice that if I replace the X in the density function for an exponential RV with the X=-(1/theta) ln(U) it reduces to theta*(U). Not sure how this helps but it seemed like too much of a coincidence. I thought if i took the expected value of an exponential(1/theta) it would equal X= -(1/theta) ln(E(U)) but E(U) is 1/2 and therefore X=(1/theta)* ln(0.5). plz help me understand how to properly approach this, i have been working on it for hours, but my teachers notes dont cover it and i cant find a reliable help online
In these cases, work with the cumulative distribution function.

$\mathbb{P}(X \leqslant t)=\mathbb{P}\left(-\tfrac 1 \theta \ln(U) \leqslant t\right)$
Now remember that U is in (0,1). So ln(U)<0. Hence -ln(U)/theta>0
So if t is $\leqslant$0, the probability is 0. So let's see for t>0

$\mathbb{P}\left(-\tfrac 1 \theta \ln(U) \leqslant t\right)=\mathbb{P}(\ln(U) \geqslant -\theta t)$
Now, remember that the exponential function is strictly increasing over the real numbers.
So $x \geqslant y \Rightarrow e^x \geqslant e^y$
Hence :
$\mathbb{P}(\ln(U) \geqslant -\theta t)=\mathbb{P}(e^{\ln(U)} \geqslant e^{-\theta t})=\mathbb{P}(U \geqslant e^{-\theta t})=1-\mathbb{P}(U \leqslant e^{-\theta t})$

Now remember that t>0. So $e^{-\theta t}<1$, since theta*t>0
And $e^{-\theta t}>0$, since an exponential is always positive.
Hence, by definition of the cumulative distribution function of a uniformly distributed rv, $\mathbb{P}(U \leqslant e^{-\theta t})=e^{-\theta t}$

And finally, we have :
$\mathbb{P}(X \leqslant t)=1-e^{-\theta t}$, if t>0.
If t<0, then the probability is 0.

This is the exact cdf of an exponential distribution !

It's referred as the "inversion formula" in my notes, but I don't know how you guys call it.

3. I just use the derivatives of the cdfs,
$f_X(x)=f_U(u)\biggl|{du\over dx}\biggr|$,
which is basically calculus one u-substitution.

4. $f_U(u)=1$ on $0.
Let $X=-\theta\ln U$
which is the same as $U=e^{-X/\theta}$.
Thus
$f_X(x)=f_U(u)\biggl|{du\over dx}\biggr|
=1\cdot \biggl|{-e^{-x/\theta}\over \theta}\biggr|
={e^{-x/\theta}\over \theta}$
.
Where $0, ${-x\over \theta} <0$, ${x\over \theta} >0$, $x>0$.

I just realized my $\theta$ is upside down, but the technique is the same, or just let $\beta={1\over\theta}$.