Sum of Xn = X1+X2+X3... Xn probability density function using convolution

Question

I am currently struggling to find the full function of the sum of a large number of probabiliy density function (PDF) for random variable. All I find online corresponds to the sum of two variables being:

$$ f_{X_1+X_2}\left(x_3\right)=\ \int^{\infty }_{-\infty }{f_{X_1}({x_3-x}_1)f_{X_2}(x_2)}dx_2\ $$

I also find some examples with more than two variables, but either using binomial or normal distribution.So what is the general function for $$ f_{X_1+X_2+X_3...X_N}\left(x_N\right) = ? $$

I've derived a potential answer with this equation below. t $$ f_{X_N}\left(x_n\right)=\int{\int{\int{\dots \int^{\infty }_{-\infty }{\prod^N_n{f_{X_{n+1}}\left(x_{n+2}-x_{n+1}\right)}dx_{n+2}}}}}\ $$

Looking at it, I'm almost certain that this is completely wrong. Any help would be apreciated!

Answer 1

Alright so to answer my own answer using the help of @mathreadler, this is what I derived:

f and g are two probability distributions satisfying. Assuming f is a probability density function and F is its representative in the frequency domain. Fourier transform of f is

$$ \mathrm{f}\left(\mathrm{x}\right)\mathrm{=\ }\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu }\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu}\mathrm{x}}\mathrm{d}\mathrm{\nu}} $$ and the inverse fourier transform is $$ \mathrm{F(}\mathrm{\nu}\mathrm{)}\mathrm{=\ }\int^{\infty }_{-\infty }{f\left(\mathrm{x}\right){\mathrm{e}}^{\mathrm{-}\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu}\mathrm{x}}\mathrm{d}\mathrm{x}}\ $$ The sum of the distribution $\mathit{\Gamma}$ is defined as the convolution of the of the two distribution so that: $$ (f*g)(x)=\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')f\ (x\ -\ x}\mathrm{')}\mathrm{dx'}} $$ $$ =\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')}\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu}\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu }\mathrm{(}\mathrm{x}\mathrm{-}{\mathrm{x}}^{\mathrm{'}}\mathrm{)}}\mathrm{d}\mathrm{\nu }}\mathrm{dx'}} $$ $$ =\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')}\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu}\right)\frac{{\mathrm{e}}^{\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}\mathrm{x}}}{{\mathrm{e}}^{\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}{\mathrm{x}}^{\mathrm{'}}}}\mathrm{d}\mathrm{\nu}}\mathrm{dx'}} $$ $$ =\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')}\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu}\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}\mathrm{x}}{\mathrm{e}}^{\mathrm{-}\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}{\mathrm{x}}^{\mathrm{'}}}\mathrm{d}\mathrm{\nu}}\mathrm{dx'}} $$ $$ =\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')}\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu}\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}\mathrm{x}}{\mathrm{e}}^{\mathrm{-}\mathrm{2}\mathrm{\pi}\mathrm{i}\mathrm{\nu}{\mathrm{x}}^{\mathrm{'}}}\mathrm{d}\mathrm{\nu}}\mathrm{dx'}} $$ $$ =\ \int^{\infty }_{-\infty }{\mathrm{g(x\ ')}{\mathrm{e}}^{\mathrm{-}\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu}{\mathrm{x}}^{\mathrm{'}}}\mathrm{dx'}\int^{\infty }_{-\infty }{\mathrm{F}\left(\mathrm{\nu}\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu}\mathrm{x}}\mathrm{d}\mathrm{\nu}}} $$ $$ \left(f*g\right)\left(x\right)=\ \int^{\infty }_{-\infty }{G\left(\mathrm{\nu}\right)\mathrm{F}\left(\mathrm{\nu}\right){\mathrm{e}}^{\mathrm{2}\mathrm{\pi }\mathrm{i}\mathrm{\nu}\mathrm{x}}\mathrm{d}\mathrm{\nu}} $$

$$ \mathcal{F}(f*g)(x))=G(\mathrm{\nu}\mathrm{)}F(\mathrm{\nu}\mathrm{)} $$

${\mathcal{F}}^{-1}$ being the inverse Fourier transform: $$ \left(f*g\right)\left(x\right)={\mathcal{F}}^{-1}(G\left(\mathrm{\nu}\right)F\left(\mathrm{\nu}\right)) $$

So basically, get the two distribution f and g, calculate their Fourier transform. Multiply them together and get the inverse Fourier transform to get $f*g$. $$ \left(f_1*f_2*f_3\dots f_N\right)\left(x\right)=\ {{\mathcal{F}}^{-1}(F}_1\left(\mathrm{\nu }\right)F_2\left(\mathrm{\nu }\right)F_3\left(\mathrm{\nu }\right)\dots F_N\left(\mathrm{\nu}\right)) $$ $$ \left(f_1*f_2*f_3\dots f_N\right)\left(x\right)=\ {\mathcal{F}}^{-1}\left(\prod^N_{n=1}{F_N\left(\mathrm{\nu}\right)}\right) $$

Answer 2

You can avoid integrals and reduce the calculations to an elementwise product if you turn the convolutions into products by using the Fourier transform of each density function and combine with the convolution theorem.

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But if you just want to use the definition with convolutions then at least you can use the result that the convolutional product is commutative, but it will still be kind of convoluted to work with and try and simplify.