The random variable follows the standard normal distribution . Given , the random variable follows the normal distribution . (1) Determine the distribution of ; (2) Compute .

Probability Theory Solution – Problem 36: Determine the distribution of ;

Question

The random variable $X$ follows the standard normal distribution $N(0,\ 1)$ . Given $X=x$ , the random variable $Y$ follows the normal distribution $N(x,\ 1)$ . (1) Determine the distribution of $Y$ ; (2) Compute $P(X Y\geqslant0)$ .

Step-by-step solution

1. Structural decomposition of the random variable From the given condition $Y|X=x \sim N(x, 1)$ , it follows that given $X=x$ , $Y$ equals $x$ plus a standard normal random variable. We can express $Y$ structurally as: $Y = X + Z$ where $X \sim N(0, 1)$ and $Z \sim N(0, 1)$ , with $Z$ independent of $X$ . (Justification: when $X$ is fixed at $x$ , $X+Z$ follows $N(x, 1)$ , consistent with the problem statement.)

2. Computing the expectation and variance of $Y$ Since $X$ and $Z$ are independent and both normally distributed, their linear combination $Y$ is also normally distributed. Expectation: $E[Y] = E[X + Z] = E[X] + E[Z] = 0 + 0 = 0$ Variance: $Var(Y) = Var(X + Z) = Var(X) + Var(Z)$ (by independence) $Var(Y) = 1 + 1 = 2$

3. Conclusion The random variable $Y$ follows a normal distribution with mean 0 and variance 2. That is, $Y \sim N(0, 2)$ .

1. Reformulating the problem Substituting $Y = X + Z$ into the inequality $XY \geqslant 0$ : $P(XY \geqslant 0) = P(X(X + Z) \geqslant 0) = P(X^2 + XZ \geqslant 0)$

2. Geometric probability via polar coordinates Since $X \sim N(0, 1)$ , $Z \sim N(0, 1)$ , and they are independent, the joint density of the random vector $(X, Z)$ possesses rotational symmetry. The probability distribution depends only on the angle. Introduce polar coordinates in the $(X, Z)$ plane: $X = R \cos \theta$ $Z = R \sin \theta$ where $R > 0$ , $\theta \in [0, 2\pi)$ , and the probability density is uniform over $\theta$ .

3. Analyzing the angular range satisfying the condition Substituting polar coordinates into the inequality $X^2 + XZ \geqslant 0$ : $R^2 \cos^2 \theta + R^2 \cos \theta \sin \theta \geqslant 0$ Dividing by $R^2$ (since $R^2 > 0$ ): $\cos \theta (\cos \theta + \sin \theta) \geqslant 0$

We seek the intervals of $\theta$ for which the above product is nonnegative. The boundary points are $\cos \theta = 0$ (i.e., $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ ) and $\cos \theta + \sin \theta = 0$ (i.e., $\tan \theta = -1$ , giving $\theta = \frac{3\pi}{4}, \frac{7\pi}{4}$ ).

Sign analysis over $[0, 2\pi)$ : * Interval $(0, \frac{\pi}{2})$ : $\cos \theta > 0$ , $\cos \theta + \sin \theta > 0$ . Product $> 0$ . (Satisfied) Arc length: $\frac{\pi}{2}$ . * Interval $(\frac{\pi}{2}, \frac{3\pi}{4})$ : $\cos \theta < 0$ , $\sin \theta > |\cos \theta| \Rightarrow$ sum $> 0$ . Product $< 0$ . * Interval $(\frac{3\pi}{4}, \frac{3\pi}{2})$ : $\cos \theta < 0$ , $\sin \theta < |\cos \theta|$ or both negative $\Rightarrow$ sum $< 0$ . Product $> 0$ . (Satisfied) Arc length: $\frac{3\pi}{2} - \frac{3\pi}{4} = \frac{3\pi}{4}$ . * Interval $(\frac{3\pi}{2}, \frac{7\pi}{4})$ : $\cos \theta > 0$ , $\sin \theta$ negative with large absolute value $\Rightarrow$ sum $< 0$ . Product $< 0$ . * Interval $(\frac{7\pi}{4}, 2\pi)$ : $\cos \theta > 0$ , $\sin \theta$ negative with small absolute value $\Rightarrow$ sum $> 0$ . Product $> 0$ . (Satisfied) Arc length: $2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$ .

4. Computing the total probability The total arc length satisfying the condition is: $L = \frac{\pi}{2} + \frac{3\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$ Since the angle is uniformly distributed, the probability is: $P(XY \geqslant 0) = \frac{L}{2\pi} = \frac{3\pi/2}{2\pi} = \frac{3}{4}$

Final answer

(1) $Y \sim N(0, 2)$ (2) $P(XY \geqslant 0) = \frac{3}{4}$

Marking scheme

The following is the complete marking rubric for this probability theory problem.

I. Checkpoints (max 7 pts total)

Part 1: Distribution of $Y$ (3 points) [additive]

1 pt: Model construction. Write the structural decomposition $Y = X + Z$ and state that $Z$ is independent of $X$ (or $Z \sim N(0,1)$ ), OR set up the correct total probability formula / marginal density integral $f_Y(y) = \int f_{Y|X}(y|x)f_X(x)dx$ , OR write the characteristic function in product form.
1 pt: Derivation. Use independence to derive $E[Y]=0$ and $Var(Y)=2$ (the step $1+1=2$ must be shown), OR correctly complete the integration / completing-the-square computation, OR simplify the characteristic function.
1 pt: Final conclusion. Explicitly state $Y \sim N(0, 2)$ (parameters must be specified).
*Note: If $N(0,2)$ is stated without any derivation, this part receives 0 points.*

Part 2: Computing $P(XY \ge 0)$ (4 points)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Structural decomposition and polar coordinates (standard approach)
1 pt: Inequality transformation. Convert the condition $XY \ge 0$ into a form involving independent variables, such as $X(X+Z) \ge 0$ or $X^2 + XZ \ge 0$ . [additive]
2 pts: Region analysis. Correctly determine the region in the $(X, Z)$ plane satisfying the condition (e.g., using polar coordinates to obtain a total arc length of $3\pi/2$ , or correctly analyzing the range of $Z$ for $X>0$ and $X<0$ ). [additive]
1 pt: Result. Obtain the probability $3/4$ (or $0.75$ ). [additive]
Chain B: Bivariate normal geometric method (Sheppard's theorem / correlation coefficient)
1 pt: Distribution identification. Explicitly state that $(X, Y)$ follows a bivariate normal distribution. [additive]
1 pt: Correlation computation. Correctly compute the correlation coefficient $\rho = \frac{1}{\sqrt{2}}$ (or covariance $Cov(X,Y)=1$ with correct variances). [additive]
1 pt: Geometric formula application. Apply the arcsine formula $P = \frac{1}{2} + \frac{\arcsin \rho}{\pi}$ , or derive the result using the geometric angle properties of the density contour lines. [additive]
1 pt: Result. Obtain the probability $3/4$ . [additive]
Chain C: Joint density integration method
1 pt: Integral setup. Use symmetry to write $2P(X>0, Y>0)$ or set up the double integral with the correct joint density $f(x,y)$ . [additive]
2 pts: Integration. Correctly evaluate the definite integral via a change of variables (e.g., $u=x, v=y/x$ ) or conversion to polar coordinates. [additive]
1 pt: Result. Obtain the probability $3/4$ . [additive]

Total (max 7)

II. Zero-credit items

In Part 1, merely listing formulas (e.g., the normal density formula) without substituting the problem's conditions ( $Y|X$ ) or performing any computation.
In Part 1, stating $Y \sim N(0,2)$ directly without providing any justification (e.g., "by properties").
In Part 2, incorrectly assuming $X$ and $Y$ are independent, yielding the result $1/2$ .
In Part 2, incorrectly assuming $X$ and $Y$ are perfectly positively correlated (i.e., $Y=kX, k>0$ ), yielding the result $1$ .

III. Deductions

Logic gap (-1): In Part 1, using $Var(X+Z) = Var(X) + Var(Z)$ without mentioning "independence" or "zero covariance" anywhere on the paper.
Inequality handling error (-1): In Part 2, dividing both sides of $X(X+Z) \ge 0$ by $X$ without considering the reversal of the inequality sign when $X<0$ (even if the correct region is obtained fortuitously via symmetry).
Arithmetic error (-1): The approach is entirely correct, but an error occurs in simple arithmetic or trigonometric values (e.g., $\arctan(-1)$ ).
*Note: The total score after deductions cannot fall below 0; if multiple solution methods are present, only the highest-scoring one is graded, and deductions are not applied redundantly.*

Probability Theory – Problem 36: Determine the distribution of ;