Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The following problem is from Jacod's Probability Essentials: enter image description here

Let $$g(y)=-\frac{1}{\lambda}\ln{y}$$ and $$ h(y)=g^{-1}(y)=e^{-\lambda y} $$ The probability density function of $Y$ can be quickly calculated: $$ f_Y(y)=f_U(h(y))|h'(y)|, \quad y\geq 0 $$ and the proof follows.

Why on earth do we need $1-U$ and how is it supposed to be used in this exercise?

share|improve this question
2  
The standard result is that if $X$ has CDF $F(x)$, then the function $F(\cdot)$ applied to $X$ yields a random variable $Y = F(X)$ that is uniformly distributed on $(0,1)$. Going the other way, if $Y \sim U(0,1)$, then $F^{-1}(Y)$ yields $X$. For an exponential random variable with parameter $\lambda$, $F(x) = 1 - \exp(-\lambda x)$, and so $F^{-1}(y) = -\frac{1}{\lambda}\ln (1-y)$. Thus, $-\frac{1}{\lambda}\ln(1-Y)$ gives an exponential random variable. So does $-\frac{1}{\lambda}\ln(Y)$ since $1-Y \sim U(0,1)$ just as $Y \sim U(0,1)$. –  Dilip Sarwate Oct 2 '12 at 2:50
    
@DilipSarwate, Thanks. I think this is how the hint is supposed to use. –  Jack Oct 2 '12 at 15:13

1 Answer 1

up vote 3 down vote accepted

I'm not familiar with this book, but I suspect the idea is to argue with cumulative distribution functions in the following manner: Let $y>0$, then $${\rm F}_Y(y)={\rm P}(Y\leq y)={\rm P}\Bigl(-{1\over\lambda}\ln U\leq y\Bigr)={\rm P}\bigl(U\geq {\rm e}^{-\lambda y}\bigr)$$ Now, if the reader thinks it is absolutely essential to turn the last inequality sign the right way before computing the probability, it is possible to use the hint and say that the last expression is $$={\rm P}\bigl(1-U\geq {\rm e}^{-\lambda y}\bigr)={\rm P}(U\leq1-{\rm e}^{-\lambda y}\bigr)={\rm F}_U\bigl(1-{\rm e}^{-\lambda y}\bigr)=1-{\rm e}^{-\lambda y}$$ This is the desired cumulative distribution function, so $Y$ is exponentially distributed with parameter $\lambda$, as stated.

Of course, we would say that it is much more natural to use $${\rm P}\bigl(U\geq {\rm e}^{-\lambda y}\bigr)=\int_{{\rm e}^{-\lambda y}}^1{\rm d}u=1-{\rm e}^{-\lambda y}$$ but for some reason students often seem to disagree with us on that...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.