考研数学概率论题,求做

如题所述

概率密度函数p(x,y)=1/4
EU=∫∫(x+y)²/4dxdy
=(1/4)∫(0,2)dy∫(0,2)(x+y)²dx
=(1/4)∫(0,2)dy∫(0,2)(x²+2xy+y²)dx
=(1/4)∫(0,2)dy[x³/3+x²y+y²x](0,2)
=(1/4)∫(0,2)[2³/3+2²y+2y²]dy
=(1/4)∫(0,2)[8/3+4y+2y²]dy
=(1/4)[8y/3+2y²+2y³/3](0,2)
=(1/4)[8×2/3+2×2²+2×2³/3]
=(1/4)[16/3+8+16/3]
=(1/4)[56/3]
=14/3追答

根据定义做
DU=∫∫[(x十y)²-14/3]²(1/4)dxdy
=(1/4) ∫∫[x²十2xy十y²-14/3]²dxdy
= (1/4) ∫∫[x^4十4x²y²十y^4十196/9十4x³y十2x²y²-28x²/3十4xy³-56xy/3-28y²/3]dxdy
= (1/4) ∫(0,2)dx∫(0,2)[x^4十4x²y²十y^4十196/9十4x³y十2x²y²-28x²/3十4xy³-56xy/3-28y²/3]dy

=(1/4)∫(0,2)dx[x^4.y十2x²y³十y^5/5十196y/9十2x³y²-28x²y/3十xy^4-28xy²/3-28y³/9](0,2)
= (1/4)∫(0,2)[2x^4十16x²十32/5十392/9十8x³-56x²/3十16x-112x/3-224/9]dx
= (1/4)∫(0,2)[2x^4 十8x³ -8 x² /3-64x/3 十1128/45]dx
= (1/4)[2x^5/5 十2x^4-8 x³/9-32x²/3 十1128x/45] (0,2)
= (1/4)[64/5 十32-64/9-128/3 十2256/45]
= 16/5 十8-16/9-32/3 十564/45
=508/45
=11.289

=(1/4)∫(0,2)dx[x^4.y十2x²y³十y^5/5十196y/9十2x³y²-28x²y/3十xy^4-28xy²/3-28y³/9](0,2)
= (1/4)∫(0,2)[2x^4十16x²十32/5十392/9十8x³-56x²/3十16x-112x/3-224/9]dx

= (1/4)[2x^5/5 十2x^4-8 x³/9-32x²/3 十1128x/45] (0,2)
= (1/4)[64/5 十32-64/9-128/3 十2256/45]
= 16/5 十8-16/9-32/3 十564/45
=508/45
=11.289

= (1/4)[2x^5/5 十2x^4-8 x³/9-32x²/3 十1128x/45] (0,2)

= (1/4)[64/5 十32-64/9-128/3 十2256/45]
= 16/5 十8-16/9-32/3 十564/45
=508/45
=11.289

= (1/4)[64/5 十32-64/9-128/3 十2256/45]

= 16/5 十8-16/9-32/3 十564/45
=508/45
=11.289

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