MHB Computing joint cumulative distribution function

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To compute the joint cumulative distribution function (CDF) for the exponential random variable X and Y = X^3, the correct formulation is F(x,y) = P(X ≤ x, Y ≤ y) = P(X ≤ x, X^3 ≤ y) = P(X ≤ min(x, y^{1/3})). The probability density function (PDF) for X is f_x(x) = λe^{-λx} for x ≥ 0. The PDF for Y is derived as f_y(y) = (λ/3)y^{-2/3}e^{-λy^{1/3}} for y ≥ 0. The next step involves integrating the joint PDF to find the joint CDF based on the established relationship. This process is essential for understanding the behavior of the random variables involved.
vincentvance
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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...
 
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Hi,

What you've done so far is correct:
vincentvance said:
$F(x,y) = P(X ≤ x, Y ≤ y) = P(X ≤ x, X^3 ≤ y) = P(X ≤ x, X ≤ y^{1/3}) = P(X ≤ \min(x, y^{1/3}))$,

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