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 $\chi$ $\sigma$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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