Re: Marginal density of normal+beta random variable



Richard Hofler wrote:

Suppose this error term in a regression: Z = X + Y where X is distributed as normal(0, sigma^2) and 0 < Y < 1 is distributed as beta(a, b).

The zi can be estimated in each observation but xi and yi cannot be separated from each other.

I need the marginal density f(z).

I can find the joint density of Z and Y as
f(z, y) = ((1 - y)^(b -1)*y^(a - 1)*Gamma[a + b])/(E^((y + z)^2/(2*Sigma^2))*
(Sqrt[2*Pi]*Sigma*Gamma[a]*Gamma[b]))

I cannot find a way to integrate out Y from f(z, y) to give me f(z).

Can someone show me how to evaluate this integral so I can obtain a closed form result?

Alternatively, is there another approach that will give me a closed form for f(z)?



I assume, as your formula indicates, that X and Y are independent. By convolution, the marginal density f for Z is

f(z) = G(a+b)/[G(a)G(b) sqrt(2pi v)] integral(y=0..1, exp(-(z-y)^2 / (2v)) y^(a-1) (1-y)^(b-1))

where G is the gamma function and v is the variance sigma^2. You can write this as an expectation:

f(z) = G(a+b)/[G(a)G(b)] E[(sigma W+z)^(a-1) (1 - (sigma W + z))^(b-1) 1_{-z/sigma <W<(1-z)/sigma}],

where W has standard normal distribution.

I highly doubt you will be able to express this in "closed form." When a and b are integers, you can probably express this in terms of the cumulative distribution function Phi for the standard normal or, equivalently, in terms of the error function erf. This might also be able to do so when a and b are integer multiples of 1/2. For example, when sigma = a = b = 1 (i.e., the simplest possible case), f(z) = Phi(1- z) - Phi(-z).

--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey

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