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Liviu Nicolaescu
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I can write a formula, but I doubt you will like it. Set

$$ \rho(x_1,\dotsc, x_n)=\frac{1}{(2\pi)^{\frac{n}{2}}\sigma_1\cdots\sigma_n} \exp\left(-\sum\frac{(x_i-\mu_i)^2}{2\sigma_i^2}\right) $$

denote the joint probability density of $(X_1,\dotsc, X_n)$ which I assumed to be independent. The probability density of $D$ is

$$ d(t)=\frac{1}{2\sqrt{t}}\int_{|x|=\sqrt{t}} \rho(x) dS(x) $$,$$ d(t)=\frac{1}{2\sqrt{t}}\int_{|x|=\sqrt{t}} \rho(x) dS(x), $$

where $dS(x)$ denotes the area element on the sphere $\{|x|=\sqrt{t}\}$.

I can write a formula, but I doubt you will like it. Set

$$ \rho(x_1,\dotsc, x_n)=\frac{1}{(2\pi)^{\frac{n}{2}}\sigma_1\cdots\sigma_n} \exp\left(-\sum\frac{(x_i-\mu_i)^2}{2\sigma_i^2}\right) $$

denote the joint probability density of $(X_1,\dotsc, X_n)$ which I assumed to be independent. The probability density of $D$ is

$$ d(t)=\frac{1}{2\sqrt{t}}\int_{|x|=\sqrt{t}} \rho(x) dS(x) $$,

where $dS(x)$ denotes the area element on the sphere $\{|x|=\sqrt{t}\}$.

I can write a formula, but I doubt you will like it. Set

$$ \rho(x_1,\dotsc, x_n)=\frac{1}{(2\pi)^{\frac{n}{2}}\sigma_1\cdots\sigma_n} \exp\left(-\sum\frac{(x_i-\mu_i)^2}{2\sigma_i^2}\right) $$

denote the joint probability density of $(X_1,\dotsc, X_n)$ which I assumed to be independent. The probability density of $D$ is

$$ d(t)=\frac{1}{2\sqrt{t}}\int_{|x|=\sqrt{t}} \rho(x) dS(x), $$

where $dS(x)$ denotes the area element on the sphere $\{|x|=\sqrt{t}\}$.

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Liviu Nicolaescu
  • 34.7k
  • 2
  • 91
  • 165

I can write a formula, but I doubt you will like it. Set

$$ \rho(x_1,\dotsc, x_n)=\frac{1}{(2\pi)^{\frac{n}{2}}\sigma_1\cdots\sigma_n} \exp\left(-\sum\frac{(x_i-\mu_i)^2}{2\sigma_i^2}\right) $$

denote the joint probability density of $(X_1,\dotsc, X_n)$ which I assumed to be independent. The probability density of $D$ is

$$ d(t)=\frac{1}{2\sqrt{t}}\int_{|x|=\sqrt{t}} \rho(x) dS(x) $$,

where $dS(x)$ denotes the area element on the sphere $\{|x|=\sqrt{t}\}$.