Representing the Probability Distribution of XY in a Graph

Alexsandro
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Could someone help me to find the probability distribution de XY below ?
Take \Omega to be a set of 5 real numbers. Define a probability measure and a random variable X on it which takes the values 1, 2, 3, 4, 5 with probability \frac{1}{10}, \frac{1}{10}, \frac{1}{5}, \frac{1}{5}, \frac{2}{5}, respectively; another random variable Y which takes the values \sqrt{2}, \sqrt{3}, \pi with probabilities \frac{1}{5}, \frac{3}{10}, \frac{1}{2}. Find the probability distribution of XY.
 
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Well, where are you stuck? What have you done, and what do you think you need to do?
 
Just go ahead and do it! Since one possible value for x is 1 (with probability 1/10) and one possible value for y is \sqrt{2} (with probability 1/5, one possible value for xy is 1*\sqrt{2}= \sqrt{2} with probability (1/10)(1/5)= 1/50. There are 15 possible values for xy. Calculate each of them.
 
thanks

HallsofIvy said:
Just go ahead and do it! Since one possible value for x is 1 (with probability 1/10) and one possible value for y is \sqrt{2} (with probability 1/5, one possible value for xy is 1*\sqrt{2}= \sqrt{2} with probability (1/10)(1/5)= 1/50. There are 15 possible values for xy. Calculate each of them.

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Thank you, for help. One doubt, that I have, consists of knowing the best way to represent the probability distribution on this situation: for a graph or correspondence between points ?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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