Understanding Independence Conditions

We are given two random variables, $X$ and $Y$. Their marginal distributions are identical standard normal distributions, meaning $X \sim N(0, 1)$ and $Y \sim N(0, 1)$. The question asks which conditions guarantee that $X$ and $Y$ are independent.

Analysis of Option 1: Zero Covariance

The condition is $Cov(X, Y) = 0$.

• Zero covariance implies independence only if the random variables are known to be jointly normally distributed.
• The problem states $X$ and $Y$ have normal marginal distributions, but not necessarily a joint normal distribution.
• Counterexample: Let $X \sim N(0, 1)$ and $Y = XZ$, where $Z$ is a standard normal variable independent of $X$. Then $Y$ also has a marginal distribution $N(0, 1)$. We have $Cov(X, Y) = E[X(XZ)] - E[X]E[XZ] = E[X^2 Z] - 0 = E[X^2]E[Z] = 1 \times 0 = 0$. However, $X$ and $Y$ are dependent since $Y$ is a function of $X$.
• Therefore, $Cov(X, Y) = 0$ is not sufficient on its own.

Analysis of Option 2: Linear Combination Normality

The condition is that $aX + bY$ is normally distributed with mean $0$ and variance $a^2 + b^2$ for all real numbers $a$ and $b$.

• This condition implies that the random vector $\begin{pmatrix} X \\ Y \end{pmatrix}$ follows a bivariate normal distribution.
• Consider the specific case where $a=1$ and $b=1$. Then $X+Y \sim N(0, 1^2 + 1^2) = N(0, 2)$.
• The variance of $X+Y$ is calculated as $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X, Y)$. Given $Var(X)=1$ and $Var(Y)=1$, we have $Var(X+Y) = 1 + 1 + 2Cov(X, Y) = 2 + 2Cov(X, Y)$.
• Since $Var(X+Y)$ must equal $2$ (from the condition), we have $2 = 2 + 2Cov(X, Y)$, which implies $Cov(X, Y) = 0$.
• Because $\begin{pmatrix} X \\ Y \end{pmatrix}$ is bivariate normal and $Cov(X, Y) = 0$, $X$ and $Y$ are independent.
• Therefore, this condition implies independence.

Analysis of Option 3: Joint Probability Condition

The condition is $P(X \le 0, Y \le 0) = 1/4$.

• Since $X \sim N(0, 1)$, $P(X \le 0) = 1/2$. Similarly, $P(Y \le 0) = 1/2$.
• If $X$ and $Y$ were independent, then $P(X \le 0, Y \le 0) = P(X \le 0)P(Y \le 0) = (1/2)(1/2) = 1/4$. This shows the condition is necessary for independence.
• However, it is not sufficient. Consider $X \sim N(0, 1)$ and $Y = UX$, where $U$ is a random variable independent of $X$ such that $P(U=1) = 1/2$ and $P(U=-1) = 1/2$. This construction yields $Y \sim N(0, 1)$ marginally, $P(X \le 0, Y \le 0) = 1/4$, but $X$ and $Y$ are dependent.
• Therefore, this condition does not guarantee independence.

Analysis of Option 4: Characteristic Function Property

The condition is $E[e^{itX + isY}] = E[e^{itX}] E[e^{isY}]$ for all real numbers $s$ and $t$.

• $E[e^{itX + isY}]$ is the joint characteristic function of $(X, Y)$, denoted $\phi_{X,Y}(t, s)$.
• $E[e^{itX}]$ and $E[e^{isY}]$ are the marginal characteristic functions of $X$ and $Y$, denoted $\phi_X(t)$ and $\phi_Y(s)$.
• The condition $\phi_{X,Y}(t, s) = \phi_X(t) \phi_Y(s)$ is the definition of independence for random variables $X$ and $Y$.
• Therefore, this condition implies independence.

Conclusion

The conditions that imply independence between $X$ and $Y$ are Option 2 and Option 4.