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We observe the input samples as $Z = X+Y$, where distribution of $Z$ could be estimated by histogram of samples and $X,Y$ are two independent random variables. One of the variable was known following exponential distribution (as $X \sim Exp(\lambda)$).

I want to get the parameter of $X$, $\lambda$, and probablity density of $Y$ with its parameters also. How to decompose the random variable $X+Y$?

Further more, what if the distribution of $X$ is unknown?

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  • $\begingroup$ Kind of related question is addressed in this blog I wrote up recently. $\endgroup$ – broccoli Nov 18 '12 at 0:58
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The problem is ill-posed. There is no unique solution if you dont specify Z. Think about it. Pick any Y with any distribution you choose. X+Y will have some distribution that depends on what you chose for Y. So if you specify Z then Y is determined but two distinct choices for Y will give two different distributions for Z. There are infinitely many solutions to your problem.

Now given that of course your second problem is also ill-posed. Just knowing that X and Y a re independent tells you almost nothing about Z. Take X and Y normal and Z will be normal. If X and Y are identical distributed Cauchys Z will be Cauchy. Two independent chi squares lead to Z be chi-square. So there are three solutions to problem 2 and that is only scratching the surface

In problem 1 you need to know Z's distribution to determine Y's. In problem 2 you cant get Z without specifying the distributions of X and Y.

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  • $\begingroup$ I can infer distribution of $Z$ by the samples as input. But it is not a typical distribution.@MichealChernick $\endgroup$ – Readon Shaw Sep 23 '12 at 13:41
  • $\begingroup$ I dont know what you mean. You can certain fit a distribution to data if you know the family and you can construct a density nonparametrically using kernel density estimation. But that doesnt precisely give you a mathematical formula for its distribution. $\endgroup$ – Michael R. Chernick Sep 23 '12 at 14:05
  • $\begingroup$ You said that if I specify $Z$ then $Y$ is determined. I can use kernel density estimation from samples instead of $Z$. But in my case parameter $\lambda$ of $X$ is unkown. So I want to estimate the $\lambda$ and pdf of $Y$. Furthermore, parameters of $Y$ are also needed to be estimated.@MichaelChernick $\endgroup$ – Readon Shaw Sep 23 '12 at 14:30
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    $\begingroup$ @MichaelChernick: I am not sure I agree with the first part of your reply. The distribution of $Z$ is the convolution of the distribution of $Y$ by an exponential distribution. For a given $\lambda$, there should be a single correspondence between $f_Z$ and $f_X$. $\endgroup$ – Xi'an May 19 '13 at 19:19
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    $\begingroup$ @Xi'an I agree with you: the first part of this answer is wrong (-1). It implicitly confuses estimation with mathematical solution. Given just the sample, there are many possible estimands for $Y$, but this does not preclude the existence of good estimators of $Y$, perhaps obtained through deconvolution. $\endgroup$ – whuber May 19 '13 at 20:09
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As stated, the law of $Z$ is the convolution of the law of $X$ and the law of $Y$. Finding the density of $X$ is called density deconvolution and $Y$ is generally considered as additive noise. Usually such a problem is considered by taking the fourier transforms of each variable (provided they exist) such that $f^{Ft} _Z = f^{Ft} _X \times f^{Ft} _Y$, where $f_Y ^{Ft}$ is the charateristic function of an exponential distribution in your case. Even if you were willing to give a parametric law to $X$, it is unusual that an analytic expression of $Z$ can be recovered.

Provided the Fourier transform of $Y$ never vanishes you may write $f^{Ft} _X = f^{Ft} _Z / f^{Ft} _Y$. Given a sample of realizations of $Z$ and replacing $f_Z ^{Ft}$ by the empirical characteristic function, the numerical integration back from the Fourier domain is a difficult task since the integral is generally not defined on $\mathbb{R}$. There are several methods that allow to do so if you have realizations of thes r.v. and a large sample. and either truncate the integral over a reasonable domain or use an appropriate kernel function. In both cases you are stuck with the choice of something like a bandwidth parameter for which there are solutions.

I must say however that the joint estimation of $\lambda$ and $f_X$ seems very ambitious and would require to construct an optimization scheme that encompasses this parameter as well...writing this down properly might indeed turn out to be ill-posed in most cases...

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  • $\begingroup$ It is ill-posed density deconvolution problem. Can you provide some issues related with the solution of such problem in real? btw: have you ever seen any tools can help us decompose the estimated pdf to theoretical basis distribution? @julienstirnemann $\endgroup$ – Readon Shaw Oct 5 '12 at 3:30
  • $\begingroup$ sure: you should look at the "deamer" library for R...for non-parametric density deconvolution under several hypotheses. It will yield a non parametric estimate of the density you are interested but it will not estimate $\lambda$! $\endgroup$ – julien stirnemann Oct 5 '12 at 5:03

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