New answers tagged cumulative-distribution-function
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Discrete analog of CDF: "cumulative mass function"?
I think "cumulative mass function" is correct, but it hasn't been widely adopted just yet. It makes sense to me as a more specific cumulative distribution function, a sibling to probability ...
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Integral of cdf of a symmetric random variable
Words are superfluous:
... but sadly I need more than 22 characters.
- 270k
5
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Accepted
Integral of cdf of a symmetric random variable
Essentially translating whuber's comment into analysis and using point symmetry of the cdf around $(0,1/2)$, $F(k)=1-F(-k)$ or $F(-k)=1-F(k)$,
$$
\begin{align*}
\int_{-k}^{k}F(x)dx&=\int_{-k}^{0}F(...
- 29.7k
1
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How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
The general approach of using a change of variables does work, with a slight modification of the strategy:
Engineer the change of variables so that the lower bound of the integral approaches $\mu$ as ...
- 5,969
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Cumulative distribution function of mixed variables
Yes, your CDF expression is correct for $a \in \{1,2,3\}$ and $b \in [0,1]$.
In general, for discrete $X$ and continuous $Y$ you have
$$
F_{X,Y}(a,b)= \sum_{x\leq a}\int_{-\infty}^b f_{X,Y}(x,y)dy
$$
...
- 1,303
1
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Accepted
How do we define the pdf in the multi-variate case and compute expectations?
Short answer: there is no inconsistency. I want to make the following two remarks:
Merely from the differentiation perspective, your confusion is understandable. For example, consider a bivariate ...
- 10.4k
6
votes
Accepted
How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
Perhaps the simplest elementary yet rigorous proof employs the cumulant-generating functions. If you insist, you can translate this into integrals involving the distribution functions. Anything else ...
- 306k
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How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
(community wiki) Failed attempt: What we want to show is
$$ \lim_{\alpha \to \infty} \frac{\mathcal{B}\left(\frac{\alpha}{\alpha + \mu} ; \alpha, k + 1 \right)}{\mathcal{B}(\alpha, k+1)} = \frac{\...
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