3
votes
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 ...
- 49.6k
2
votes
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 ...
- 97.6k
2
votes
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, ...
- 2,156
2
votes
Accepted
The distribution density of a random variable $N=\max (X,Y)$
In general, when $X$ and $Y$ are independent and identically distributed,
$$P\{N \leq t\} = P\{X \leq t, Y \leq t\} = P\{X \leq t\} P\{Y \leq t\} = P\{X \leq t\}^2. $$
The first equality is because $\...
- 2,522
2
votes
Distribution of random variable $Y = (X + 1)^2$.
$$ P (Y \leq u) = P((X+1)^2 \leq u) = P(X \leq \sqrt{u}-1)$$
Thus, using $F$ for the cummulative distribution function we have
$$F_Y (u) = F_x (\sqrt{u}-1)$$
The density $f$ is a derivative of $F$:
$$...
- 1,033
2
votes
Accepted
Related to order statistics?
Using the assumptions we can develop the equation you provided.
Let
$T_1 = \min_{i}{Z_i}$, and let's find the PDF of $T_1$, by first deriving the CDF of $T_1$:
$$P(T_1 \leq t) = P(\min_{i}{Z_i} \leq t)...
- 102
1
vote
Related to order statistics?
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}\,\...
- 120k
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