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
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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 $F(-k)=1-F(k)$,
$$
\begin{align*}
\int_{-k}^{k}F(x)dx&=\int_{-k}^{0}F(...
- 29.7k
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How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
The general approach of using a change of variables does work, with a slight modification of the strategy:
Engineer the change of variables so that the lower bound of the integral approaches $\mu$ as ...
- 5,969
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Cumulative distribution function of mixed variables
Yes, your CDF expression is correct for $a \in \{1,2,3\}$ and $b \in [0,1]$.
In general, for discrete $X$ and continuous $Y$ you have
$$
F_{X,Y}(a,b)= \sum_{x\leq a}\int_{-\infty}^b f_{X,Y}(x,y)dy
$$
...
- 1,303
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Accepted
How do we define the pdf in the multi-variate case and compute expectations?
Short answer: there is no inconsistency. I want to make the following two remarks:
Merely from the differentiation perspective, your confusion is understandable. For example, consider a bivariate ...
- 10.4k
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Accepted
How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
Perhaps the simplest elementary yet rigorous proof employs the cumulant-generating functions. If you insist, you can translate this into integrals involving the distribution functions. Anything else ...
- 306k
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How to show that the negative binomial CDF converges to the Poisson CDF? (Incomplete beta vs incomplete gamma functions)
(community wiki) Failed attempt: What we want to show is
$$ \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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