Let X and Y be two independent random variables. Which one of the relations between expectation (E), variance (Var) and covariance (Cov) given below is FALSE?

Understanding Independent Random Variables

In probability theory and statistics, random variables are considered independent if the outcome of one does not affect the outcome of the other. This property of independence simplifies many calculations involving their expectations, variances, and covariances. The question asks us to identify the statement that is FALSE among the given relations for two independent random variables X and Y.

Analyzing Properties of Independent Variables

Let's examine each given relation one by one, keeping in mind that X and Y are independent random variables.

- **Expectation of Product: $E(XY)=E(X)E(Y)$ This is a fundamental property of independent random variables. If X and Y are independent, the expected value of their product is indeed the product of their individual expected values. This is a very useful simplification. Therefore, this relation is TRUE.
- Covariance of Independent Variables: $Cov(X,Y)=0$ ** The covariance between two random variables X and Y is defined as: $$Cov(X,Y)=E(XY)-E(X)E(Y)$$ Since X and Y are independent, we know from the previous point that $E(XY)=E(X)E(Y)$. Substituting this into the covariance formula: $$Cov(X,Y)=E(X)E(Y)-E(X)E(Y)=0$$ Thus, if two random variables are independent, their covariance is always zero. It's important to note that the converse is not always true; zero covariance does not necessarily imply independence. Therefore, this relation is TRUE.
- **Variance of Sum: $Var(X+Y)=Var(X)+Var(Y)$ ** The general formula for the variance of the sum of two random variables is: $$Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)$$ Because X and Y are independent, we've established that $Cov(X,Y)=0$. Substituting this into the formula: $$Var(X+Y)=Var(X)+Var(Y)+2(0)$$ $$Var(X+Y)=Var(X)+Var(Y)$$ This property is often called the additivity of variance for independent variables. Therefore, this relation is TRUE.
- **Expectation of Squared Product: $E(X^2{Y}^2)=(E(X){)}^2(E(Y){)}^2$ ** Let's analyze this relation: - If X and Y are independent, then any functions of X and Y, say $g(X)$ and $h(Y)$, are also independent. In this case, $X^2$ and $Y^2$ are independent. - Since $X^2$ and $Y^2$ are independent, we can apply the property of expectation of a product of independent variables: $$E(X^2{Y}^2)=E(X^2)E(Y^2)$$ - Now, let's consider the right side of the given relation: $(E(X){)}^2(E(Y){)}^2$. - We know that for a general random variable A, $E(A^2)$ is not necessarily equal to $(E(A){)}^2$. In fact, the variance is defined as $Var(A)=E(A^2)-(E(A){)}^2$, which implies $E(A^2)=Var(A)+(E(A){)}^2$. - Unless $Var(X)=0$ and $Var(Y)=0$ (which means X and Y are constants, not truly random variables in a meaningful sense), $E(X^2)$ is generally greater than $(E(X){)}^2$, and $E(Y^2)$ is generally greater than $(E(Y){)}^2$. - Therefore, $E(X^2)E(Y^2)$ is generally not equal to $(E(X){)}^2(E(Y){)}^2$. Consider an example: Let X be a random variable that takes values -1 and 1 with equal probability (0.5 each). Then $E(X)=(-1)(0.5)+(1)(0.5)=0$. And $X^2$ always takes the value 1. So, $E(X^2)=1$. In this case, $E(X^2)=1$ but $(E(X){)}^2=0^2=0$. Clearly, $E(X^2)\ne (E(X){)}^2$. Thus, the relation $E(X^2{Y}^2)=(E(X){)}^2(E(Y){)}^2$ is generally FALSE.

Conclusion

Based on the analysis, the relation that is FALSE for two independent random variables X and Y is $E(X^2{Y}^2)=(E(X){)}^2(E(Y){)}^2$.