การกระจายตัวของความแปรปรวนของตัวแปรกำลังสองและไคสแควร์?

ปัญหาต่อไปนี้เกิดขึ้นเมื่อเร็ว ๆ นี้ขณะวิเคราะห์ข้อมูล หากตัวแปรสุ่ม X ตามการแจกแจงปกติและ Y ตามการแจกแจง $\chi^2_n$ (ด้วย n dof) $Z = X^2 + Y^2$ กระจายอย่างไร ถึงตอนนี้ฉันมากับ pdf ของ $Y^2$ :

\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}

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

เช่นเดียวกับการทำให้เข้าใจง่ายสำหรับอินทิกรัลรวม ( มี pdf กับ m dof): $X^2$ $\chi^2_m$

\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
แหล่งที่มา

If the

Y

$Y$ wasn't squared, I'd have some specific advice. I don't think this one will be tractable (nor necessarily particularly enlightening even if it were to prove tractable). I'd be tempted to look at computational approaches, like numerical convolution or simulation, depending on exactly what you want to do with the result.

— Glen_b -Reinstate Monica

It's very unlikely in my opinion that the integral can be done.

— Dave31415

@ Dave31415 สำหรับ

และ

แม้หนึ่งสามารถคำนวณอย่างชัดเจนสำหรับค่าหนึ่งในเชิงบวกของ

และม.

มันจะเท่ากับการรวมกันเชิงเส้นของฟังก์ชันเลขชี้กำลังและข้อผิดพลาดกับสัมประสิทธิ์ที่เป็นพหุนามใน

n

$n$

m

$m$

n

$n$

m

$m$

. การประเมินผลสามารถดำเนินการผ่านการเปลี่ยนตัวผู้เล่น

ตัวอย่างเช่นด้วย

เราได้รับ

\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

In case it helps, the variable $Y^2$ 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
แหล่งที่มา

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
แหล่งที่มา

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