Solve p = P(2X <= Y^2) using double integral

[Addez123]

Homework Statement

p = P(2X <= Y^2) where,
X = Y =
F(x) = 1 - 1/x^3, x >= 1
F(x) = 0, x < 1

Calculate p using a double intergral.

Relevant Equations

Pareto distribution

Background information
Earlier they've shown that some double integrals can be simulated if it contains pdfs.
Ex: $$\int \int cos(xy)e^{-x-y^2} dx dy$$
Can be solved by setting:
Exponential distribution
$$f(x) = e^{-x}, Exp(1)$$
Normal distribution
$$f(y) = e^{-y^2}, N(0, 1/\sqrt 2)$$By knowing this we can simulate a lot of $$N(0, 1/\sqrt 2)$$ and Exp(1) values, and put them into $$cos(xy)$$ then take the mean value of all those products to get the mean, aka. the most probable answer.

Problem
In this exercise I'm suppose to solve the probability using double integrals and I have no clue where to start.
I can integrate both X and Y but what's the point?
Am I suppose to solve
$$ \int _1^\infty \int _1^\infty 2X - Y^2 dx dy$$

That's not going to give me a value between 0 and 1.
This is where I'm stuck.

 

[anuttarasammyak]

Homework Statement:: p = P(2X <= Y^2) where,
X = Y =
F(x) = 1 - 1/x^3, x >= 1
F(x) = 0, x < 1

Calculate p using a double intergral.

I could not understand what it means. Could you write down the formula you are struggling with?

 

[Addez123]

X and Y both use the same cdf, which is F(x)
The cdf F(x) is 1 - 1/x^3 when x >= 1 and 0 otherwise.

My struggle is how am I suppose to write the double integral that solves the probability p.

 

[anuttarasammyak]

Thanks. Though I am not accustomed to Pareto distribution, how about
$$\int_1^\infty \frac{3}{x^4}dx \int_1^\infty \frac{3}{y^4}dy\ \theta(y^2-2x)$$
$$=\int_1^\infty \frac{3}{x^4}dx \int_{\sqrt{2x}}^\infty \frac{3}{y^4}dy\$$
where $\theta(t)=1$ for t>0, 0 for t<0.

 

[BvU]

Do you understand that a reader can not even come to a problem description on the basis of what you post in #1 ?

Yet you bring in 'backgound information' that may or may not be useful once the problem statement is completed.

Start with deriving the pdf for $X$ and $Y$, then put into words what $P(2x \le y^2)$ means.

The double integral should contain probability distributions, not the condition (that goes into the bounds) !

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