1

Let $X$ and $Y$ be i.i.d. Expo(λ), and $T = \log(X/Y )$. Find the CDF and PDF of $T$.

$P(T \le t)=P \left(\log(\frac XY) \le t\right)=P(X \le 10^tY)$

\begin{align} P(T \le t) & = P(X \le 10^tY)=P(X \le 10^tY \mid Y=y)P(Y=y) \\[10pt] & =(1-e^{λ10^ty})λe^{-λy} = λe^{-λy}-λe^{-λy(10^{ty}+1)} \end{align}

$$f(t)=-λ^2e^{-λy}-λ^2(10^{ty}+1)e^{-λy(10^{ty}+1)}; \quad y \ge 0$$

Is it wrong?

Michael Hardy
  • 1
user13
  • 1,661
  • 1
    I wondered how the number $10$ gets involved. Then I remembered that some people construe "$\log$" as meaning $\log_{10}.$ But almost certainly, whoever wrote this intended $\log_e. \qquad$ – Michael Hardy Dec 21 '17 at 18:22
  • 1
    $P(Y=y)$ is zero, so that can't be right when you write the CDF of $T$. This should be replaced with the density of $Y$, and you have to integrate the entire expression over $y$. – StubbornAtom Dec 21 '17 at 18:48
  • @StubbornAtom Thanks – user13 Dec 21 '17 at 18:49

0 Answers0