MHB Computing joint cumulative distribution function

[vincentvance]

With X having the exponential $(\lambda)$ distribution and $Y = X^3$, how do I compute the joint cumulative distribution function?

Here is how far I've come:

$F(x,y) = P(X ≤ x, Y ≤ y) = P(X ≤ x, x^3 ≤ y) = P(X ≤ x, X ≤ y^{1/3}) = P(X ≤ min(x, y^{1/3})$,

$f_x(x) = \lambda e^{-\lambda x}$$ for $ x ≥ 0, 0 otherwise,

$f_y(y) = \lambda /3 y^{-2/3}e^{-\lambda y^{1/3}}$$ for $ y ≥ 0, 0 otherwise.

Now I have no idea how to continue...

 

[Siron]

Hi,

What you've done so far is correct:

$F(x,y) = P(X ≤ x, Y ≤ y) = P(X ≤ x, X^3 ≤ y) = P(X ≤ x, X ≤ y^{1/3}) = P(X ≤ \min(x, y^{1/3}))$,

We know that $X$ has an exponential distribution with parameter $\lambda$, hence the joint distribution is ...