Timeline for Does convolution of a probability distribution with itself converge to its mean?
Current License: CC BY-SA 4.0
15 events
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| Feb 14, 2022 at 9:20 | comment | added | Emma | @MichaelHardy it is the weighted average of i.i.d random variables. The answer provided below is a more accurate description of my problem and I agree with his answer. | |
| Feb 14, 2022 at 5:30 | vote | accept | Emma | ||
| Feb 14, 2022 at 1:30 | comment | added | Iosif Pinelis | Do you have a response to the comments and answer on this page? | |
| Feb 10, 2022 at 23:26 | history | became hot network question | |||
| S Feb 10, 2022 at 22:26 | history | edited | Glorfindel | CC BY-SA 4.0 |
typos corrected
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| S Feb 10, 2022 at 22:26 | history | suggested | dodo | CC BY-SA 4.0 |
clarification
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| Feb 10, 2022 at 18:28 | answer | added | Iosif Pinelis | timeline score: 13 | |
| Feb 10, 2022 at 17:54 | comment | added | Michael Hardy | I wonder whether what you want is a weighted average of i.i.d. random variables, rather than assigning weights to the density functions? | |
| Feb 10, 2022 at 17:24 | comment | added | Michael Hardy | If $f$ is a probability density function then the convolution $f*f$ is also a probability density function, so if $0<\lambda<1$ is a scalar, then then $\lambda^2 f*f$ and $(1-\lambda)^2 f*f$ and $\lambda(1-\lambda)f*f$ are functions whose integrals are less than $1.$ and $f*f$ is the density function of a distribution whose variance is twice that of the distribution whose density is $f.$ So this is moving away from the delta function, not toward it. | |
| Feb 10, 2022 at 17:09 | review | Suggested edits | |||
| S Feb 10, 2022 at 22:26 | |||||
| Feb 10, 2022 at 17:08 | comment | added | dodo | @IosifPinelis (iii) I think by "converging to the mean", she means that if we repeat the convolution operation, then the distribution will finally become a Dirac-delta distribution at the mean. (ii) Convolution means this" en.wikipedia.org/wiki/Convolution_of_probability_distributions. (i) $f$ usually means the probability density function of a distribution, I think. | |
| Feb 10, 2022 at 16:34 | comment | added | Iosif Pinelis | (i) What do you mean by "a probability distribution $f(x)$"? A probability distribution is a measure. (ii) What do you mean by $\lambda f$? If $f$ is indeed a probability distribution and $\lambda f$ is the usual product, then the convolutions will converge to the zero measure (even in total variation). (iii) How can "the resulting distribution converge[] to the mean"? Again, a probability distribution is a measure, whereas the mean is number. | |
| Feb 10, 2022 at 16:23 | history | edited | YCor |
edited tags
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| S Feb 10, 2022 at 15:26 | review | First questions | |||
| Feb 10, 2022 at 15:34 | |||||
| S Feb 10, 2022 at 15:26 | history | asked | Emma | CC BY-SA 4.0 |