Exercise
ABy Admin
Jun 25'24
Answer
Solution: E
We have
[[math]]
E((X-Y)^2]=E[X^2] + E[Y^2] - 2E[XY]=E((X+Y)^2] = E[X^2] + E[Y^2] + 2E[XY].
[[/math]]
Hence [math]E[XY] = 0 [/math]. Then we also have
[[math]]
E[X^2] = E[(X-Y)^2] = E[X^2] + E[Y^2] = 2E[X^2].
[[/math]]
Hence [math]E[X^2] = 0 [/math] which means that [math]P(X=0) = 1 [/math].