MHB Proving of Y=g(X) as a continuous random variable

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If X is a continuous random variable and g is a continuous function defined on X, then Y = g(X) is also a continuous random variable. The proof relies on the continuity of the cumulative distribution function F_X(x), which is continuous for continuous random variables. Since g is continuous, the inverse function g^{-1}(y) is also continuous, allowing for the transformation of probabilities. Consequently, the cumulative distribution function F_Y(y) can be expressed in terms of F_X, confirming that Y maintains continuity. This establishes that Y = g(X) is indeed a continuous random variable.
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If X is a continuous random variable and g is a continuous function
defined on X (Ω), then Y = g(X ) is a continuous random variable.
Prove or disprove it.
 
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bl00d said:
If X is a continuous random variable and g is a continuous function
defined on X (Ω), then Y = g(X ) is a continuous random variable.
Prove or disprove it.

If X is a continuous r.v. then $F_{X} (x) = P \{ X < x\}$ is continous. Now if $y=g(x)$ is continuous then $x=g^{-1} (y)$ is also continuous and the same is for... $$F_{Y} (y) = P \{g(X) < y\} = P \{X < g^{-1} (Y)\} = F_{X} (g^{-1} (y))$$Kind regards $\chi$ $\sigma$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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