特定Beta分布随机变量X+Y的分布证明及PDF求解技术问询

Alright, let's break down how to derive the probability density function (PDF) of (Z = X + Y) here—you've got solid experimental observations pointing to (Z \sim B(1, M-1)), so let's formalize that with actionable, targeted steps:

1. Start by Rewriting (X) and (Y) Using Gaussian Vector Structure #

Since (X) and (Y) are dependent, we can't just convolve their individual Beta PDFs directly. Instead, lean into the underlying Gaussian vector setup:

• Given (V \perp W) and (|V|^2 = |W|^2 = C) (a constant, since their norms are fixed), simplify (X = \frac{(U^T V)^2}{C |U|^2}) and (Y = \frac{(U^T W)^2}{C |U|^2}).
• Construct an orthonormal basis for (\mathbb{R}^{2M}) that includes (\frac{V}{|V|}) and (\frac{W}{|W|}). Decompose (U) into components along this basis and the orthogonal subspace:
  [
  U = a \cdot \frac{V}{|V|} + b \cdot \frac{W}{|W|} + U_\perp
  ]
  Here, (a) and (b) are independent zero-mean Gaussians, and (U_\perp) is orthogonal to both (V) and (W) (so it lives in a (2M-2)-dimensional subspace).
• Substitute into (X) and (Y):
  [
  X = \frac{a^2}{a2 + b^2 + S}, \quad Y = \frac{b^2}{a2 + b^2 + S}
  ]
  where (S = |U_\perp|^2) (a chi-squared random variable with (2(M-1)) degrees of freedom).

2. Isolate (Z = X + Y) and Use Independent Chi-Squared Ratios #

Notice (Z = X + Y = \frac{a^2 + b^2}{a2 + b^2 + S}). Let's define:

• (T = a^2 + b^2): this is a chi-squared variable with 2 degrees of freedom (sum of squares of two independent Gaussians), so its PDF is (f_T(t) = \frac{1}{2}e^{-t/2}) for (t > 0).
• (S) is independent of (T) (since (U_\perp) is independent of (a,b)), with PDF (f_S(s) = \frac{s^{M-2}e{-s/2}}{2^{M-1}\Gamma(M-1)}) for (s > 0).

For independent non-negative random variables (T) and (S), the PDF of (Z = \frac{T}{T+S}) can be derived via variable transformation:

• Let (z = \frac{t}{t+s}), which rearranges to (s = t \cdot \frac{1-z}{z}). The Jacobian of this transformation is (\frac{t}{z^2}).
• Substitute into the joint PDF integral:
  [
  f_Z(z) = \int_0^\infty f_T(t) \cdot f_S\left(t \cdot \frac{1-z}{z}\right) \cdot \frac{t}{z^2} dt
  ]
  Plug in the chi-squared PDFs and simplify the integral—you'll find it evaluates exactly to the PDF of a (B(1, M-1)) distribution.

3. Use a Standard Beta Distribution Shortcut #

There's a well-known result here: if (Z = \frac{T}{T+S}) where (T \sim \chi^2(2k)) and (S \sim \chi^2(2l)) are independent, then (Z \sim B(k, l)). In your case:

• (T) is (\chi^2(2)) (so (k=1))
• (S) is (\chi^2(2(M-1))) (so (l=M-1))

This directly gives (Z \sim B(1, M-1)). The PDF of this Beta distribution simplifies to:
[
f_Z(z) = (M-1)(1-z)^{M-2} \quad \text{for } 0 < z < 1
]
Which you can cross-verify with the integral from step 2.

4. Confirm with Moment Generating Functions (MGFs) #

To double-check, compute the MGF of (Z) and show it matches the MGF of (B(1, M-1)):

• The MGF of (B(\alpha, \beta)) is (M_Z(t) = {}_1F_1(\alpha; \alpha+\beta; t)). For (\alpha=1), (\beta=M-1), this becomes (M_Z(t) = \frac{M-1}{M-1 - t}) (valid for (t < M-1)).
• Compute the MGF of (Z = \frac{T}{T+S}) using independence of (T) and (S):
  [
  M_Z(t) = \mathbb{E}\left[e^{t \cdot \frac{T}{T+S}}\right] = \int_0^\infty \int_0^\infty e^{t \cdot \frac{t'}{t'+s'}} f_T(t') f_S(s') dt' ds'
  ]
  Evaluating this integral will confirm it matches the Beta MGF, solidifying your conclusion.

内容的提问来源于stack exchange，提问作者ishmam zabir