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Thread: Confused about rand. var. transforms

  1. #1
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    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
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  2. #2
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    Hello,
    Quote Originally Posted by nine4fours View Post
    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.


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


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

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


    And finally, we have :
    $\displaystyle \mathbb{P}(X \leqslant t)=1-e^{-\theta t}$, 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.
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  3. #3
    MHF Contributor matheagle's Avatar
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    I just use the derivatives of the cdfs,
    $\displaystyle f_X(x)=f_U(u)\biggl|{du\over dx}\biggr|$,
    which is basically calculus one u-substitution.
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  4. #4
    MHF Contributor matheagle's Avatar
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    $\displaystyle f_U(u)=1$ on $\displaystyle 0<u<1$.
    Let $\displaystyle X=-\theta\ln U$
    which is the same as $\displaystyle U=e^{-X/\theta}$.
    Thus
    $\displaystyle f_X(x)=f_U(u)\biggl|{du\over dx}\biggr|
    =1\cdot \biggl|{-e^{-x/\theta}\over \theta}\biggr|
    ={e^{-x/\theta}\over \theta}$.
    Where $\displaystyle 0<e^{-x/\theta}<1$, $\displaystyle {-x\over \theta} <0$, $\displaystyle {x\over \theta} >0$, $\displaystyle x>0$.

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