Joint Probability density function

[Guest]

f(X,Y)=(2/3)(x^2)(e^-xy) 1<x<a, 0<y<[tex]\infty[/tex]
1. Calculate the value of a.
2. Calculate the probability that X is less than 1.75 and Y is greater than 2.

[HallsofIvy]

Did you make any attempt to do this yourself? It's pretty straight forward if you know the basic definitions.

[tex]\frac{2}{3}\int_{x=1}^a\int_{y= 0}^\infty x^2 e^{-xy} dy dx= 1[/tex]
Do the integration and solve for a.

[tex]\frac{2}{3}\int_{x= 1}^{1.75}\int_{y=2}^\infty x^2 e^{-xy} dy dx[/tex].

[Guest]

[tex]f(x,y)= \frac{2}{3} x^{2 } e^{-xy }[/tex]

[tex]x \in (1,a)[/tex]

[tex]a>1[/tex]

[tex]y \in (0, \infty )[/tex]

[tex]\int\limits_{0}^{ \infty } \int\limits_{1}^{a}f(x,y)dxdy=1[/tex]

[tex]\int\limits_{0}^{ \infty } \int\limits_{1}^{a} \frac{2}{3} x^{2 } e^{-xy }dxdy=\frac{2}{3}\int\limits_{0}^{ \infty } \int\limits_{1}^{ a } x^{2 } e^{-xy }dx dy=\frac{2}{3}\int\limits_{1}^{ a } \int\limits_{0}^{ \infty } x^{2 } e^{-xy }dy dx = \frac{2}{3}\int\limits_{1}^{ a } x^{2 } \int\limits_{0}^{ \infty } e^{-xy }dy dx = \frac{2}{3}\int\limits_{1}^{ a } x^{2 } \frac{-1}{x} e^{-xy } |_{0 } ^{ \infty } dx=\frac{2}{3}\int\limits_{1}^{ a } x (- e^{-xy }) |_{0 } ^{ \infty } dx=\frac{2}{3}\int\limits_{1}^{ a } x dx=\frac{2}{3} \frac{1}{2} x^{2 } |_{1 } ^{a }=\frac{1}{3} x^{2 } |_{1 } ^{a }=\frac{1}{3} (a^{2 }-1)[/tex]

[tex]\frac{1}{3} (a^{2 }-1)=1[/tex]

[tex]a^{2 }-1=3[/tex]

[tex]a^{2 }=4[/tex]

[tex]a= \sqrt{4}[/tex]

[tex]a= 2[/tex]

[tex]P(X<1,75,Y>2))=\int\limits_{2}^{ \infty } \int\limits_{1}^{1,75}f(x,y)dxdy=\int\limits_{2}^{ \infty } \int\limits_{1}^{1,75} \frac{2}{3} x^{2 } e^{-xy }dxdy=\frac{2}{3}\int\limits_{2}^{ \infty } \int\limits_{1}^{ 1,75 } x^{2 } e^{-xy }dx dy=\frac{2}{3}\int\limits_{1}^{ 1,75 } \int\limits_{2}^{ \infty } x^{2 } e^{-xy }dy dx = \frac{2}{3}\int\limits_{1}^{ 1,75 } x^{2 } \int\limits_{2}^{ \infty } e^{-xy }dy dx = \frac{2}{3}\int\limits_{1}^{ 1,75 } x^{2 } \frac{-1}{x} e^{-xy } |_{2 } ^{ \infty } dx=\frac{2}{3}\int\limits_{1}^{ 1,75 } x (- e^{-xy }) |_{2 } ^{ \infty } dx=\frac{2}{3}\int\limits_{1}^{ 1,75 } x e^{-2x } dx=\frac{2}{3}( -\frac{1}{2}x e^{-2x }-\frac{1}{4}e^{-2x } )|_{1 } ^{ 1,75}=-\frac{1}{6}(2x e^{-2x }+e^{-2x })|_{1 } ^{ 1,75}=-\frac{1}{6}(3,5 e^{-3,5 }+e^{-3,5 }-2 e^{-2 }-e^{-2 })=-\frac{1}{6}(4,5 e^{-3,5 }-3 e^{-2 })=0,5e^{-2 }-0,75 e^{-3,5 } \approx 0,045[/tex]