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 The joint density function of the random variables X and Y is \text{ The joint density function of the random variables } X \text{ and } Y \text{ is }

f(x,y)=6x, for 0<x<1, 0<y<1x,f(x,y)=6x, \text{ for } 0<x<1, \text{ } 0<y<1-x,

f(x,y)=0, elsewhere.f(x,y)=0, \text{ elsewhere.}

[

(a) Show that X and Y are not independent. (a) \text{ Show that } X \text{ and } Y \text{ are not independent. }

(b) Find P(X>0.3Y=0.5).(b) \text{ Find } P(X >0.3 | Y = 0.5).

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(a)\textbf{(a)} The variables XX and YY will be independent\text{\underline{independent}} if the equality

f(x,y)=g(x)h(y)\begin{align} f(x,y)=g(x)h(y) \end{align}

holds for all\textbf{all} values xx and yy which the variables are able to assume.

In order to check the independence of the variables, let us find their marginal densities.

g(x)=yf(x,y)dy=01x6xdy=(6xy)01x=6x((1x)0)=6x(1-x), for 0<x<1\begin{align*} g(x) &= \int_y f(x,y)dy \\ &= \int_0^{1-x} 6xdy \\ &= \bigr( 6xy \bigr) \biggr |_0^{1-x} \\ &= 6x\bigr( (1-x) - 0 \bigr) \\ &= \textbf{6x(1-x)} \text{, for } 0<x<1 \end{align*}

Further on, we need to find h(y)h(y). For some arbitrary y<0,1>y \in <0,1>, because of y<1xy<1-x we also have:

x<1y  0<x<1yx<1-y \text{ } \Rightarrow \text{ } 0<x<1-y

Thus, similarly as for g(x)g(x), we obtain:

h(y)=xf(x,y)dx=01y6xdx=(3x2)01y=3((1y)202)=3(1-y)2, for 0<y<1\begin{align*} h(y) &= \int_x f(x,y)dx \\ &= \int_0^{1-y} 6xdx \\ &= \bigr( 3x^2 \bigr) \biggr |_0^{1-y} \\ &= 3\bigr( (1-y)^2 - 0^2 \bigr) \\ &= \textbf{3(1-y)}^{\textbf{2}} \text{, for } 0<y<1 \end{align*}

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