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Given $X\sim \exp(\lambda)$ and $Y = X^3$, what would be their joint cumulative distribution, $F(x, y)$? Since $X$ and $Y$ are dependent, I can't just integrate the product of their probability density functions: $f(x)f(y)$. I've got it down to:

$\begin{align*} F(x,y) &= P(X \le x, Y \le y) \\ &= P(X \le x, X^3 \le y) \\ &= P(X \le x, X \le y^{1/3}) \\ &= P(X \le \min(x, y^{1/3})) \end{align*}$

but don't know how to proceed from here on. I've determined the probability density functions for f(x) and f(y). If they were independent, I would just integrate f(y)f(x) dx dy Since they are not independent, I'm not sure how to derive their joint probability distribution. Once I have that, my assumption is that I can just integrate that to derive their joint cdf.

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  • $\begingroup$ The joint CDF needs to be specified for all $(x,y)$. It is obviously $0$ in the second, third and fourth quadrants. For the first quadrant, try finding $F_{X,Y}(x_0,y)$ for fixed $x_0 > 0$ as $y$ varies from $0$ to $\infty$. $\endgroup$
    – Dilip Sarwate
    Oct 13, 2013 at 20:26
  • $\begingroup$ @dfeuer Thanks for re-editing the question, which made no sense just a minute or so ago! $\endgroup$
    – Dilip Sarwate
    Oct 13, 2013 at 20:27
  • $\begingroup$ @DilipSarwate, I felt responsible! I voted to approve the edit and then immediately realized it was bogus. $\endgroup$
    – dfeuer
    Oct 13, 2013 at 20:29
  • $\begingroup$ Hi Dilip,thank you for your response. I've determined the individual probability density functions for X and Y. Specifically, my difficulty is with: $\endgroup$
    – EggHead
    Oct 14, 2013 at 2:47
  • $\begingroup$ 1) figuring out how to derive the joint pdf $\endgroup$
    – EggHead
    Oct 14, 2013 at 2:47

1 Answer 1

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Hint: Break it up into cases. Case 1, $0 \leq x \leq y^{1/3}$. Case 2, $0 \leq y^{1/3} < x$.

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