excans:8b16e9c194: Difference between revisions
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(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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<math display = "block"> | <math display = "block"> | ||
E[X^2] = E[(X-Y)^2] = E[X^2] + E[Y^2]. | E[X^2] = E[(X-Y)^2] = E[X^2] + E[Y^2] = 2E[X^2]. | ||
</math> | </math> | ||
Hence <math>E[X^2] = 0 </math> which means that <math>P(X=0) = 1 </math>. | Hence <math>E[X^2] = 0 </math> which means that <math>P(X=0) = 1 </math>. | ||
Latest revision as of 01:55, 25 June 2024
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].