(1) The random vector follows a normal distribution. For all , . Compute the density function of . (2) The random vector follows a normal distribution with mean zero, and . Compute

Probability Theory Solution – Problem 49: Compute the density function of

Question

(1) The random vector $\mathbf{Y}=(Y_{1},\ \cdots\ ,\ Y_{n})^{T}$ follows a normal distribution. For all $k\neq j,\,E(Y_{k}\mid Y_{j})=0$ , $E(Y_{k}^{2}\mid Y_{j})=1$ . Compute the density function of $\mathbf{Y}$ . (2) The random vector $\pmb{X}=(X_{1},\ X_{2},\ X_{3})^{T}$ follows a normal distribution with mean zero, and $E(X_{1}^{2}\mid X_{2})=X_{3}^{2}$ .

Compute $E{\frac{X_{1}^{2}+X_{2}^{2}}{X_{2}^{2}+X_{3}^{2}}}.$

Step-by-step solution

(1) For $k \neq j$ , $E(Y_k \mid Y_j) = 0$ implies that the covariance is 0 under zero mean, hence $Y_k$ and $Y_j$ are uncorrelated. Under normality, uncorrelatedness implies mutual independence. $E(Y_k^2 \mid Y_j) = 1$ implies, by independence, $E(Y_k^2) = 1$ . Therefore $Y_1,\dots,Y_n$ are i.i.d. $N(0,1)$ . The density function is: $f(\mathbf{y}) = (2\pi)^{-n/2} e^{-\frac12 \sum_{i=1}^n y_i^2}.$ (2) Given $E(X_1^2 \mid X_2) = X_3^2$ . For a normal vector, $E(X_1 \mid X_2) = a X_2$ , $\mathrm{Var}(X_1 \mid X_2) = v$ (a constant). Thus $E(X_1^2 \mid X_2) = v + a^2 X_2^2.$ Comparing with $X_3^2$ (as a function of $X_2$ ), we must have $v = 0$ and $a^2 X_2^2 = X_3^2$ , which gives $X_3 = \pm a X_2$ and $X_1 = a X_2$ . Then $X_1^2 + X_2^2 = a^2 X_2^2 + X_2^2 = (1+a^2) X_2^2$ , $X_2^2 + X_3^2 = X_2^2 + a^2 X_2^2 = (1+a^2) X_2^2$ . Hence the ratio equals 1 (almost surely). $E\left[ \frac{X_1^2 + X_2^2}{X_2^2 + X_3^2} \right] = 1.$

Final answer

(1) Density function: $f(\mathbf{y}) = (2\pi)^{-n/2} e^{-\frac12 \sum_{i=1}^n y_i^2}.$ (2) $E\left[ \frac{X_1^2 + X_2^2}{X_2^2 + X_3^2} \right] = 1.$

Marking scheme

The following is the grading rubric based on the official solution.

1. Checkpoints (max 7 pts total)

Part 1: Finding the density function of $\mathbf{Y}$ (3 pts)

Deriving zero mean and independence [1 pt]
Using $E(Y_k \mid Y_j)=0$ to show that $Y_k, Y_j$ have zero mean and are uncorrelated, then combining with the normality property to conclude mutual independence.
*Note: If the student does not mention "uncorrelated" or the derivation of "zero mean" and merely asserts standard normal distribution, no credit for this item.*
Deriving unit variance [1 pt]
Using independence and $E(Y_k^2 \mid Y_j)=1$ to obtain $E(Y_k^2)=1$ , hence $\text{Var}(Y_k)=1$ .
Writing the correct joint density function [1 pt]
Must write the product-form density of the $n$ -dimensional standard normal distribution.

Part 2: Computing the expectation (4 pts)

Establishing the structural formula for the conditional expectation [1 pt]
Using properties of the normal distribution to write $E(X_1^2 \mid X_2) = \text{Var}(X_1 \mid X_2) + [E(X_1 \mid X_2)]^2 = v + a^2 X_2^2$ (where $v, a$ are constants).
*Note: Equivalent forms such as $C_1 + C_2 X_2^2$ are accepted.*
Arguing that the conditional variance is zero [1 pt]
From the relation $v + a^2 X_2^2 = X_3^2$ , arguing that $v=0$ must hold (the constant term vanishes).
*Justification: From the range of $X_3^2$ (which includes 0) or from the functional relationship between variables.*
Determining the functional relationship among the random variables [1 pt]
Explicitly obtaining that $X_1^2$ and $X_3^2$ are the same multiple of $X_2^2$ (i.e., $X_1^2 = a^2 X_2^2$ and $X_3^2 = a^2 X_2^2$ , almost surely).
*Note: This item rewards recognizing the key step that the distribution degenerates to a singular distribution.*
Computing the final result [1 pt]
Substituting into the ratio to obtain the constant 1, and concluding that the expectation equals 1.

Total (max 7)

2. Zero-credit items

(Part 1) Merely writing down the normal distribution formula without deriving the parameters from the given conditions $E(Y_k \mid Y_j)=0$ .
(Part 2) No derivation at all; directly guessing the answer is 1.
(Part 2) Incorrectly assuming $X_1, X_2, X_3$ are mutually independent, leading to a logical contradiction (e.g., concluding that $X_3$ is a constant) yet continuing the computation.

3. Deductions

Special-case method penalty (Cap at 5/7):
If the student does not carry out a general derivation but instead directly assumes specific random variables (e.g., $X_1=X_2=X_3$ ) satisfying the conditions and computes the result, Part 2 receives at most 2 pts (only the structural formula point and the result point; logical derivation points are withheld).
Logical gap (-1):
In Part 1, failing to state that "under normality, uncorrelatedness is equivalent to independence" and jumping directly from uncorrelatedness to $E(Y_k^2 \mid Y_j)=E(Y_k^2)$ .
Notation or definition error (-1):
Omitting the domain of the density function (e.g., $\mathbf{y} \in \mathbb{R}^n$ ) is generally not penalized, but writing the vector in scalar form causing ambiguity incurs a 1-point deduction.

Probability Theory – Problem 49: Compute the density function of