Revision as of 01:48, 25 June 2024 by Admin (Created page with "'''Solution: E''' We have <math display = "block"> 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 display = "block"> E[X^2] = E[(X-Y)^2] = E[X^2] + E[Y^2]. </math> Hence <math>E[X^2] = 0 </math> which means that <math>P(X=0) = 1 </math>.")
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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]. [[/math]]

Hence [math]E[X^2] = 0 [/math] which means that [math]P(X=0) = 1 [/math].

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