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Let's say that x is a random variable which follows a conditional probability density function $f(x|N)$, where $N$ is an integer parameter. If the functional form of $f(x|N=1)$ is known, then one can calculate the convolution of this $f(x|N=1)$ with itself like the following:

$\int_{0}^{a}f(y|N=1)f(a-y|N=n-1)dy$.

Is this equal to the probability density function of $x$ given $N=n$, so f(x|N=n)? If so, could you please show a proof or explain why the two terms are the same?

For the people who wonder where the formula was seen, this is a paper showing such a formula. See equation 5.

Nownuri
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  • [https://en.wikipedia.org/wiki/Convolution_of_probability_distributions](https://en.wikipedia.org/wiki/Convolution_of_probability_distributions) – J. Delaney Feb 17 '22 at 13:32
  • Please elaborate on "what does it mean." You give an explicit formula, so what kind of answer are you seeking beyond that? Perhaps the ones at https://stats.stackexchange.com/questions/331973 are appropriate? – whuber Feb 17 '22 at 15:32
  • this is the sum of independent random variables I believe- eg https://www.statlect.com/fundamentals-of-probability/sums-of-independent-random-variables. – seanv507 Feb 17 '22 at 18:22
  • from OP's paper: "The signal distribution for an integer number of muons is given by the signal distribution of a single particle convolved multiple times with itself." – seanv507 Feb 17 '22 at 18:23
  • @whuber The OP is asking why convoluting the conditional pdf's of $x|N=n-1$ and $x|N=1$ yields the pdf of $x|N=n$. Which is not the same as asking why convoluting the pdf's of $X$ and $Y$ yields the pdf of $X+Y$ as adressed in the post you link to. – StratosFair Feb 19 '22 at 23:01
  • @StratosFair I don't see why not, but it of course depends on how $f$ is defined. I don't want to read a paper in order to determine what exactly this question is trying to ask: the duty to form a clear question that stands on it own rests with the original poser. – whuber Feb 20 '22 at 13:01

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