# Tag Info

### A uniform bound of Hölder class-like densities.

All you need is the Holder condition on $f$, not on any of its derivatives. For an unbounded nonnegative function to have integral $1$, its graph would have to be extremely steep at points. The Holder ...
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### Why does the expected value of given PDF not exist?

Your assertion $$\frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{u} \; \mathrm{d}u = \frac{ln(\lvert u \rvert)}{2\pi} \Big|_{-\infty}^{\infty}\\$$ is not correct. In fact, these four integrals all ...
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### Finding the probability density function (pdf) for a two variable 'function'

Unfortunately there is no general method that can give you $Z$ from $X,Y$. If there is a variable transformation $f$ that is bijective, and such, invertable, then there does exist a formula. However, ...
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Accepted

• 102
1 vote

Lack of independence makes this too complicated. Independent but not identically distributed is still possible to analyse: The CDF of the minimum is $$F_{Z_{(1)}}(x) = 1- \prod\limits_i (1-F_{Z_i}(z))... • 143k 1 vote ### Probability in terms of CDF and PDF Well, yes but that's quite trivial, since the first factor inside the integral is constant and what you get is just 1 times it. What you could do, is put the indicator of the event that's inside the ... • 6,934 1 vote Accepted ### PDF of a joint probability function with transformation X = Y_1 - Y_2 You have the integral trees backwards. Using your notation, you should have.$$\begin{align}\mathsf P(X\leq x) &= \int_{y_2=0}^{y_2\to\infty}\left.\int_{y_1=y_2}^{y_1=y_2+x}\mathrm e^{-y_1}\,\...
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