If [tex]x[/tex] and [tex]y[/tex] are in [tex]L¹[/tex], show that [tex]xy[/tex] is not in [tex]L¹[/tex] if [tex]x[/tex] and [tex]y[/tex] are not independent.

a) [tex]xy[/tex] is always in [tex]L¹[/tex] regardless of the independence of [tex]x[/tex] and [tex]y[/tex].

b) If [tex]x[/tex] and [tex]y[/tex] are not independent, then [tex]xy[/tex] is in [tex]L¹[/tex].

c) If [tex]x[/tex] and [tex]y[/tex] are not independent, then [tex]xy[/tex] is not in [tex]L¹[/tex].

d) The independence of [tex]x[/tex] and [tex]y[/tex] does not affect whether [tex]xy[/tex] is in [tex]L¹[/tex].

The statement in option (c) is correct. If ( x ) and ( y ) are not independent, it implies that their joint probability distribution is not simply the product of their individual probability distributions. The definition of L^1 norm for a random variable Z is given by:

||Z||_{L^1} = ∫ |Z| dP

where dP is the probability measure. For the product of two random variables X and Y, XY, to be in L^1, the expectation of its absolute value must be finite. If X and Y are not independent, the joint distribution of XY may have dependencies that affect the integrability of |XY| under the probability measure. Consequently, XY may not be in L^1 in such cases.

In other words, the lack of independence between X and Y can introduce correlations or dependencies that impact the integrability condition for |XY| to be in L^1. This highlights the importance of independence in ensuring the product of random variables lies in the L^1 space.

Therefore, option (c) is the correct answer, indicating that the independence of X and Y plays a crucial role in determining whether XY is in L^1.