Hi could anyone help me here. I am currently studying Probability and Statistics and I have just been introduced to the idea of Joint density functions. So for the question this is how far I have got:
$$f_{X,Y}(x,y)=\begin{cases} 8xy& \text{if } 0< y < x < 1\\ 0 &\text{otherwise}, \end{cases}$$
I know I need to find out firstly $(f_x|y)$ and $(f_y|x)$
With $(f_x|y)(x|y) = \frac{(f_(x,y)*(x,y)}{f_y*(y))} = \frac{8xy}{?}$
here I'm not sure what should be on the bottom. I think maybe 8x?
and then secondly I need to find out secondly $(f_y|x)$
with $(f_y|x)(y|x) = \frac{(f_(x,y)*(x,y)}{f_x*(x))} = \frac{8xy}{?}$
and again I'm not quite sure what should be on the bottom. I think maybe 8y?
Any help on this would really be appreciated. Thank you.
the bivariate support is
$$0<\underbrace{y<x<1}_{X-Support}$$
Thus to derive $f_Y(y)$ you have to integrate the joint density over $X-$ support
$$f_Y(y)=\int_y^1 8xy dx=\dots=4y(1-y^2)\cdot\mathbb{1}_{(0;1)}(y)$$
Thus the conditional density is
$$f_{X|Y}(x|y)=\frac{2x}{1-y^2}\mathbb{1}_{(y;1)}(x)$$
Thus, given that $Y=y=\frac{1}{3}$ you get
$$f_{X|Y=\frac{1}{3}}(t)=\frac{9}{4} t\cdot \mathbb{1}_{(\frac{1}{3};1)}(t)$$
Concluding,
The probability to have
$$\mathbb{P}\Bigg[X>\frac{3}{4}\Bigg|Y=\frac{1}{3}\Bigg]=\int_{\frac{3}{4}}^{1}\frac{9}{4} tdt=\frac{63}{128}\approx0.49$$