Probability- Transformation of variable

jack1234
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Following is the
question
http://tinyurl.com/62uxof
solution
http://tinyurl.com/6m4lcc

The distribution in question means cumulative distribution.

What I do not understand in the solution is the step from P{X_1<=wX_2} to the integration formula that followed immediately. May I know the reasoning for it?
 
Last edited:
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jack1234 said:
What I do not understand in the solution is the step from P{X_1<=wX_2} to the integration formula that followed immediately. May I know the reasoning for it?

Hi jack1234! :smile:

P(X ≤ wY) = lim ∑P(X ≤ wY | v ≤ Y ≤ v + dv)P(v ≤ Y ≤ v + dv)

= ∫P(X ≤ wv)dP(Y ≤ v) :wink:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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