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Let $S$ and $T$ two random variables with exponential distribution of rate $\lambda$ and density $f(u)=\lambda e^{-\lambda u},u>0$. Find the density of:

  • 1) $X=|S-T|$.

$\rightarrow X\sim Exp(\lambda)$

  • 2) $Y=S^3$.

$\rightarrow f_Y(y)=\frac{1}{3}\lambda y^{-\frac{2}3{}}e^{-\lambda y^{\frac{1}{3}}}$

  • 3) $Z=\min(S^3,T)$.

In this case I am having difficulty because, if I know that $S \perp T$, I don't know anything about the relationship between $S^3$ and $T$. Moreover, the graph doesn't help since $y=s^3$ is a function with a point of inflection in $(0,0)$. Can you help me?

Francesco Totti
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    If $S$ and $T$ are independent then so are $S^3$ and $T$. Can you continue? – StubbornAtom May 25 '20 at 14:52
  • @StubbornAtom Thanks for your answer. Well, if there was, I would have $F_{Z}(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(\min(S^3,T)\leq z)=\mathbb{P}(S^3\leq z, T\leq z)=1-\mathbb{P}(S^3>z, T>z)=1-\mathbb{P}(S^3>z)\mathbb{P}(T>z)=1-\mathbb{P}(S>z^{\frac{1}{3}})\mathbb{P}(T>z)=1-\int_{z^{\frac{1}{3}}}^{+\infty}\lambda e^{-\lambda s}ds \int_{z}^{+\infty} \lambda e^{-\lambda t}dt$ – Francesco Totti May 25 '20 at 14:58
  • @StubbornAtom But could you please justify why, if $S \perp T$, so $S^3 \perp T$? – Francesco Totti May 25 '20 at 15:00
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    You don't even need to notice that. Independence of $S$ and $T$ is enough to say $P(S^3>z,T>z)=P(S>z^{1/3},T>z)=P(S>z^{1/3})P(T>z)$ for $z>0$. – StubbornAtom May 25 '20 at 15:13
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    Some information on functions of independent random variables (noticing that $S^3$ is a function of $S$): https://math.stackexchange.com/questions/8742/are-functions-of-independent-variables-also-independent – jeremy909 May 25 '20 at 15:24

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