正方变量和卡方变量平方的卷积分布？

最近在分析数据时出现了以下问题。如果随机变量X遵循正态分布且Y遵循 $\chi^2_n$ 分布（其中n自由度），如何是 $Z = X^2 + Y^2$ 分布？到现在为止我想出的PDF $Y^2$ ：

\begin{array}{rcl} ψ_{n}^{2} (x) & = & \frac{\partial F (\sqrt{x})}{\partial x} \\ = & {(\int_{0}^{\sqrt{x}} \frac{t^{n / 2 - 1} \cdot e^{- t / 2}}{2^{n / 2} Γ (n / 2)} d t)}_{x}^{'} \\ = & \frac{1}{2^{n / 2} Γ (n / 2)} \cdot {(\sqrt{x})}^{n / 2 - 1} \cdot e^{- \sqrt{x} / 2} \cdot {(\sqrt{x})}_{x}^{'} \\ = & \frac{1}{2^{n / 2 - 1} Γ (n / 2)} \cdot x^{n / 4 - 1} \cdot e^{- \sqrt{x} / 2} \end{array}

$\begin{eqnarray} \psi^2_n(x) &=& \frac{\partial F(\sqrt{x})}{\partial x} \\ &=& \left( \int_0^{\sqrt{x}} \frac{t^{n/2-1}\cdot e^{-t/2}}{2^{n/2}\Gamma(n/2)} \mathrm{d}t \right)^\prime_x \\ &=& \frac{1}{2^{n/2}\Gamma(n/2)} \cdot \left( \sqrt{x} \right)^{n/2-1} \cdot e^{-\sqrt{x}/2} \cdot \left( \sqrt{x} \right)^\prime_x \\ &=& \frac{1}{2^{n/2-1}\Gamma(n/2)} \cdot x^{n/4-1} \cdot e^{-\sqrt{x}/2} \end{eqnarray}$

以及一些简化的卷积积分（具有PDF ，其中m自由度）： $X^2$ $\chi^2_m$

\begin{array}{rcl} K_{m n} (t) & := & (χ_{m}^{2} * ψ_{n}^{2}) (t) \\ = & \int_{0}^{t} χ_{m}^{2} (x) \cdot ψ_{n}^{2} (t - x) d x \\ = & {(2^{\frac{(n + m)}{2} + 1} Γ (\frac{m}{2}) Γ (\frac{n}{2}))}^{- 1} \cdot \int_{0}^{t} (t - x)^{\frac{n}{4} - 1} \cdot x^{\frac{m}{2} - 1} \cdot \exp (- (\sqrt{t - x} + x) / 2) d x \end{array}

$\begin{eqnarray} K_{mn}(t) &:=& ( \chi^2_m \ast \psi^2_n )(t) \\ &=& \int_0^t \chi^2_m(x) \cdot \psi^2_n(t-x) \mathrm{d}x \\ &=& \left( 2^{\frac{(n+m)}{2}+1} \Gamma(\frac{m}{2}) \Gamma(\frac{n}{2}) \right)^{-1} \cdot \int_0^t (t-x)^{\frac{n}{4}-1} \cdot x^{\frac{m}{2}-1} \cdot \exp(-(\sqrt{t-x}+x)/2) \mathrm{d}x \end{eqnarray}$

Does someone see a good way of calculating this integral for any real t or does it have to be computed numerically? Or am I missing a much simpler solution?

— Leo Szilard
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如果

不平方，我会提供一些具体建议。我认为这一点不会很容易处理（即使证明很容易处理，也不一定特别有启发性）。我很想看看计算方法，例如数值卷积或仿真，具体取决于您要对结果执行什么操作。

Y

$Y$

— Glen_b-恢复莫妮卡2014年

在我看来，积分是不可能完成的。

— Dave31415 2014年

@ Dave31415 即使对于

和

，也可以为

和

正整数值显式计算积分

n

$n$

m

$m$

n

$n$

m

$m$ 。它将等于指数和误差函数的线性组合，其系数是

中的多项式

。可以通过替换

进行评估。例如，对于

我们得到

\sqrt{t}

$\sqrt{t}$

x = t - u^{2}

$x=t-u^2$

n = 2, m = 4

$n=2,m=4$

\frac{1}{4} e^{- \frac{1}{8} (2 \sqrt{t} + 1)^{2}} (e^{\frac{\sqrt{t}}{2}} (\sqrt{2 π} (4 t + 3) (erfi (\frac{2 \sqrt{t} - 1}{2 \sqrt{2}}) + erfi (\frac{1}{2 \sqrt{2}})) + 4 e^{\frac{1}{8}}) - 4 e^{\frac{t}{2} + \frac{1}{8}} (2 \sqrt{t} + 1))

$\frac{1}{4}e^{-\frac{1}{8}(2\sqrt{t}+1)^2}(e^{\frac{\sqrt{t}}{2}}(\sqrt{2\pi} (4t+3)(\text{erfi}(\frac{2\sqrt{t}-1}{2\sqrt{2}})+\text{erfi}(\frac{1}{2\sqrt{2}}))+4e^\frac{1}{8})-4 e^{\frac{t}{2}+\frac{1}{8}}(2\sqrt{t}+1))$ .

— whuber

Nice. For odd numbers, you could probably approximate it with the average of the result for bounding even numbers? Or maybe not.

— Dave31415

Thanks for your replies! For some even-even cases I got a similar result involving Dawson's function, but it looks like I'll have to do some more work for a general solution...

— Leo Szilard

$Y^2$ is a generalised gamma random variable (see e.g., Stacy 1962). Your question is asking for the distribution of the sum of a chi-squared random variable and a generalised gamma random variable. To my knowledge, the density of the resultant variable has no closed form expression. Hence, the convolution you have obtained is an integral with no closed form solution. I think you're going to be stuck with a numerical solution for this one.

Stacy, E.W. (1962). A Generalization of the Gamma Distribution. Annals of Mathematical Statistics 33(3), pp. 1187-1192.

— Reinstate Monica
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This is a hint only. Pearson type III can be Chi-squared. Sometimes a convolution can be found by convolving something with itself. I managed to do this for convolving ND and GD, for which I convolved a Pearson III with itself. How this works with ND $^2$ and Chi-Squared, I am not sure. But, you asked for hints, and this is a general hint. That should be enough to get you started, I hope.

— Carl
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Could you explain how this answers the question? It doesn't seem directly related.

— whuber

Pearson type III convolution with itself can be done. For some reason convolving one thing with itself is easier to solve than convolving one thing with another. For example, I solved the convolution of Pearson type III and obtained the convolutions of ND with GD, a related problem.

— Carl

Doesn't seem to have helped, will delete shortly.

— Carl