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the question asks for the density of the smaller of $X$ and $Y^3$, $X$ and $Y$ both being exponentially distributed independent random variables with densities $ae^{-ax}$. I think I know that I have to start by finding $P(X = x, Y^3 > x) + P(Y^3 = x, X>x)$, and then integrate from $x$ to $\infty$. I'm not sure where to go from here.

Thanks!

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Care to accept one answer for a change? –  Did May 2 '11 at 8:38
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3 Answers 3

Hint: $P(\min(X,Y)\le x)=1-P(\min(X,Y)>x)$. $P(\min(X,Y)>x)=P(X>x,Y>x)$.

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Let $W$ be the minimum of $X$ and $Y^3$. Then $W \ge w$ iff $X \ge w$ and $Y^3\ge w$, that is, iff $X \ge w$ and $Y\ge w^{1/3}$.

The probability that $X \ge w$ is $e^{-aw}$. The probability that $Y \ge w^{1/3}$ is $e^{-aw^{1/3}}$. Multiply to get the probability that $W \ge w$. Now the cdf of $W$ is easy. Differentiate to get the density.

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Thanks so much everyone! When you say finding the cdf from W >= w is easy, do you mean you can just 1- P (w>=w). Will the >= be a problem? –  user9485 Apr 12 '11 at 11:40
    
@QeD, for absolutely continuous distribution functions (such as exponential) no it will not. The easy explanation of that is that $P(W=w)=0$, since $P(W=w)=\int_w^wp(x)dx$, where $p$ is probability density function of $W$ –  mpiktas Apr 12 '11 at 11:51
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Let $W = Y^3$. Then $$P(W \leq w) = P(Y^3 \leq w) = P(Y \leq \sqrt[3]{w})$$

$$= 1-e^{a(\sqrt[3]{w})}$$

and then use the fact $P(\min(X_1, X_2, \dots, X_n) > x) = P(X_1 >x \ \text{and} \ X_2 > x \dots)$. This is known as order statistics.

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No, this is not. –  Did Apr 12 '11 at 13:25
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