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Consider the probability triple $([0,1]^2, \mathcal{B}[0,1], Leb_2)$.

Let $X: [0,1]^2 \rightarrow \mathbb{R}$ be random variables defined by

$X(\omega_1, \omega_2) = \omega_1 + \omega_2$,

$Y(\omega_1, \omega_2) = \omega_2 - \omega_1$.

Let $F_{X,Y}$ be the joint cumulative distribution function. Find $F_{X,Y}(0.58,0.35)$.

I have attempted to solve the problem by:

$F_{X,Y}(0.58,0.35)=\mathbb{P}(X \le 0.58, Y \le 0.35)$

$=\mathbb{P}( \omega_1 + \omega_2 \le 0.58, \omega_2 - \omega_1 \le 0.35)$

$=\mathbb{P}( \omega_1 \le 0.115, \omega_2 \le 0.4625)$

$=Leb_2([0,0.115]X[0,0.4625])$

$=0.0535$

However I do not think this is correct.

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  • $\begingroup$ Please see edit of my attempt at this question. $\endgroup$ Commented Oct 18, 2021 at 13:52

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I am unsure of how I would go about finding the joint cumulative distribution function for X and Y.

If you are unsure this means that you have some ideas to develop... I suggest you to apply a change of variable, calculate the jacobian and find

$$f_{XY}(x,y)=0.5$$

in the following bivariate support

enter image description here

Thus the Joint CDF can be easily derived by integration od the pdf

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  • $\begingroup$ Hi, thanks for your response. We have not been taught the Jacobian in class yet. I had a quick look online and saw that you calculate the partial deviates of each random variable and find the determinant of the matrix of partial derivatives. Doing this using my current random variables gives a determinant of 2. I think this is why you suggested a change of variable first? $\endgroup$ Commented Oct 18, 2021 at 12:47

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