1. ## proof uniform distribution

2. Originally Posted by maths_123
1. Let $X = U + 1$.

The cdf of X is given by

$F(x) = \Pr(X < x) = \Pr(U + 1 < x) = \Pr(U < x - 1) = \int_{0}^{x - 1} 1 \, du = \, ....$

Therefore the pdf of X is given by $f(x) = \frac{dF}{dx} = \, ....$ for $1 \leq x \leq 2$ and zero elsewhere.

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2. Let $Y = U^2$.

The cdf of Y is given by

$G(y) = \Pr(Y < y) = \Pr(U^2 < y) = \Pr(-\sqrt{y} < U < \sqrt{y}) = \int_{0}^{\sqrt{y}} 1 \, du = \, ....$

Therefore the pdf of Y is given by $g(y) = \frac{dG}{dy} = \, ....$ for $0 \leq y \leq 1$ and zero elsewhere.

3. Originally Posted by mr fantastic
1. Let $X = U + 1$.

The cdf of X is given by

$F(x) = \Pr(X < x) = \Pr(U + 1 < x) = \Pr(U < x - 1) = \int_{0}^{x - 1} 1 \, du = \, ....$

Therefore the pdf of X is given by $f(x) = \frac{dF}{dx} = \, ....$ for $1 \leq x \leq 2$ and zero elsewhere.

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2. Let $Y = U^2$.

The cdf of Y is given by

$G(y) = \Pr(Y < y) = \Pr(U^2 < y) = \Pr(-\sqrt{y} < U < \sqrt{y}) = \int_{0}^{\sqrt{y}} 1 \, du = \, ....$

Therefore the pdf of Y is given by $g(y) = \frac{dG}{dy} = \, ....$ for $0 \leq y \leq 1$ and zero elsewhere.

Thanks for the help, but i still don't understand how this shows the U+1 is a uniform distribution but u^2 does not.

4. Originally Posted by maths_123
Thanks for the help, but i still don't understand how this shows the U+1 is a uniform distribution but u^2 does not.
Did you do the integrations that I left you to get the cdf in each case? Did you differentiate the cdf in each case to get the pdf? If you've done these things then it should be obvious.

Please post all that you've done in response to my first post.