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Let $X$ be uniformly distributed on $[a,b]$. Find a function $f$ such that $Y=f(X)$ is exponentially distributed with parameter $1$. How do I do this?

user91500
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user109707
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2 Answers2

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You need $\Pr(f(X)\le y) = 1-e^{-y}$ for $y>0$. That is true if $f$ is an increasing function and $\Pr(X\le f^{-1}(y))=1-e^{-y}$. But you know that $\Pr(X\le x)=(x-a)/(b-a)$ if $a\le x\le b$, so $\Pr(X\le f^{-1}(y))=(f^{-1}(y)-a)/(b-a)$. Thus it is enough to find a fuunction $f$ for which $$ \frac{f^{-1}(y)-a}{b-a} = 1-e^{-y}\text{ for }y\ge 0. $$ Can you take it from there?

Michael Hardy
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  • Is it true if $f$ is decreasing? – mookid Jun 09 '14 at 12:48
  • @mookid : Actually there are both increasing and decreasing functions, and non-monotonic functions that can serve, because instead of using the fact that $\Pr(X\le x)=(x-z)/(b-a)$, we could have used the fact that $\Pr(X\ge \text{a certain number})=(x-a)/(b-a)$, or that $\Pr(X\in\text{a certain other set})=(x-a)/(b-a)$. – Michael Hardy Jun 09 '14 at 12:57
  • my point being, if $f$ is not increasing, there is no equivalence between $f(X)\le y$ and $X\le f^ {-1}(y)$ – mookid Jun 09 '14 at 12:58
  • OK, I've revised my answer. For convenience, it now rules out non-increasing functions that would also serve. – Michael Hardy Jun 09 '14 at 13:22
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Hint: you know that $$ a<x<b\implies P(X>x) = \frac{b-x}{b-a} $$and you want $$ P(f(X)>y) = e^{-y}$$

solution: assuming that $f$ is one to one and increasing, $$ e^{-y}=P(f(X)>y) = P(X > f^{-1}(y)) = \frac{b-f^{-1}(y)}{b-a} \\ f^{-1}(y) = b-(b-a)e^{-y} $$ And solve for $y=f(x)$: $$ x=f^{-1}(y) = b-(b-a)e^{-y}\\ f(x) = y=-\log\frac{b-x}{b-a} $$

Eventually, check that $f$ is one to one and increasing.

mookid
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