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A common problem in statistics is, "Given that X is distributed according to the pdf f(x) and Y is distributed according to the pdf g(y), what is the pdf h(z) of Z = X + Y?" The answer is the convolution of f(x) and g(y):
$$ h(z)=(f*g)(z)=\int_{-\infty}^\infty f(z-t)g(t) dt = \int_{-\infty}^\infty f(t)g(z-t) dt $$
In my research I have stumbled across the reverse problem given that X and Y are IID (identically and independently distributed): "Given that the random variable Z = X+Y has a known pdf h(z), what is the pdf f(x) if it is known that X and Y are IID?"
Does anyone have insight to how this problem could be solved? Is there such a thing as a deconvolution operator that could help me? Thanks in advance.
$$ h(z)=(f*g)(z)=\int_{-\infty}^\infty f(z-t)g(t) dt = \int_{-\infty}^\infty f(t)g(z-t) dt $$
In my research I have stumbled across the reverse problem given that X and Y are IID (identically and independently distributed): "Given that the random variable Z = X+Y has a known pdf h(z), what is the pdf f(x) if it is known that X and Y are IID?"
Does anyone have insight to how this problem could be solved? Is there such a thing as a deconvolution operator that could help me? Thanks in advance.