Task
We have depth of random vector: $$f(x,y) =\begin{cases}\frac{1}{5} & \text{for } 0<x<1\;,2<y<7\\0 & \text{for other } x,y\end{cases}$$
a) Calculate shore distributions
b) Correlation coefficient for X,Y
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a)
$$f_X(x) =\begin{cases} \int_0^1 \frac{1}{5} \, dy= \frac{1}{5} & \text{for } 0<x<1 \\0 & \text{for other} \end{cases}$$
$f_Y(y)$ will also be $\frac{1}{5}$
b) Correlation coefficient is: $$p(X,Y)= \frac{\operatorname{cov}(X,Y)}{ \sqrt{D^2 X}\sqrt{D^2 Y} } $$ $EX=(EX,EY) = (\frac{1}{5},\frac{1}{5})$
what I don't know how to calculate in this task:
$EXY=\text{?} \; (\frac{1}{25}?)$
$D^2 X=\text{?}$
$D^2 Y=\text{?}$
Any help would be appreciated.
Your calculations in (a) are incorrect. Correct is $$f_X(x) =\begin{cases} \int_2^7 \frac{1}{5}\mathop{dy}= 1 & \text{if } 0<x<1 \\0 & \text{else} \end{cases}.$$ For $Y$ we find that $$f_Y(y) =\begin{cases} \int_0^1 \frac{1}{5}\mathop{dx}= \frac{1}{5} & \text{if } 2<y<7 \\0 & \text{else} \end{cases}.$$
Your expected values are incorrect as well. For example, $$\mathbb E X = \int_0^1 x f_X(x) \mathop{dx} = \int_0^1 x \mathop{dx} = \frac{x^2}{2}\Bigr|_{x=0}^1 = \frac{1}{2}.$$ The variance is $$\operatorname{Var}X = \mathbb E (X^2) - (\mathbb EX)^2 = \int_0^1 x^2 f_X(x)\mathop{dx} - \frac{1}{4} = \int_0^1 x^2 \mathop{dx} - \frac{1}{4}= \frac{x^3}{3}\Bigr|_{x=0}^1 -\frac{1}{4}= \frac{1}{3}-\frac{1}{4} = \frac{1}{12}.$$ I leave the analogous calculations for $Y$ up to you.
Moreover, $$\mathbb E (XY) = \int_0^1\int_2^7 xyf(x,y) \mathop{dy} \mathop{dx}=\int_0^1\int_2^7 \frac{xy}{5} \mathop{dy} \mathop{dx} = \int_0^1\frac{xy^2}{10}\Bigr|_{y=2}^7 \mathop{dx} =\int_0^1\frac{9x}{2} \mathop{dx} = \frac{9}{4}.$$
If you succeeded in finding $\mathbb E Y$, you can calculate the covariance using $\operatorname{Cov}(X,Y) = \mathbb E(XY) - \mathbb E X\mathbb E Y$.