Math  /  Calculus

Questionpago 34
Solution. (a) On a fX,Y(x,y)={10xy2,0xy10, ailleurs f_{X, Y}(x, y)=\left\{\begin{array}{c} 10 x y^{2}, 0 \leq x \leq y \leq 1 \\ 0, \text { ailleurs } \end{array}\right.  Pour 0<x<y<1, on a FX,Y(x,y)=0xuy10uv2dvdu=100xuuyv2 dv du=100xu(v33)uy du=100xu(y3u33)du=1030xy3uu4 du=103(y3u22u55)0x=103(y3x22x55)=53x2y323x5\begin{array}{l} \text { Pour } 0<x<y<1, \text { on a } \\ \begin{aligned} F_{X, Y}(x, y) & =\int_{0}^{x} \int_{u}^{y} 10 u v^{2} d v d u=10 \int_{0}^{x} u \int_{u}^{y} v^{2} \mathrm{~d} v \mathrm{~d} u \\ & =\left.10 \int_{0}^{x} u\left(\frac{v^{3}}{3}\right)\right|_{u} ^{y} \mathrm{~d} u=10 \int_{0}^{x} u\left(\frac{y^{3}-u^{3}}{3}\right) \mathrm{d} u \\ & =\frac{10}{3} \int_{0}^{x} y^{3} u-u^{4} \mathrm{~d} u=\left.\frac{10}{3}\left(y^{3} \frac{u^{2}}{2}-\frac{u^{5}}{5}\right)\right|_{0} ^{x} \\ & =\frac{10}{3}\left(y^{3} \frac{x^{2}}{2}-\frac{x^{5}}{5}\right) \\ & =\frac{5}{3} x^{2} y^{3}-\frac{2}{3} x^{5} \end{aligned} \end{array}

Studdy Solution
Donc, la fonction de répartition conjointe est : FX,Y(x,y)=53x2y323x5 F_{X,Y}(x,y) = \frac{5}{3}x^2y^3 - \frac{2}{3}x^5 pour 0xy10 \leq x \leq y \leq 1.

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