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Let's say we want to compute the expectancy (E(X)) of the following joint distribution:

$f(x,y)=3x,$
$_{0<y<x<1}$

I could derive some approach for solving this problem, depicted below:

A) $∫_0^x ∫_y^1 3x dxdy== x-\frac{x^4}{4}$

B) $∫_y^1∫_0^x3x^2 dydx=-\frac{3}{4}(y^4-1)$

C) $∫_0^1 ∫_0^1 3x^2 dxdy=∫_0^1 ∫_0^1 3x^2 dydx=1$

D) $∫_0^1 ∫_y^1 3x^2 dxdy=\frac{3}{4}$

All of the above seem right but I can't tell which one is the real answer and why the others are false, the book I'm reading used the D approach.
Any hint or answer would be appreciated.

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The probability density function is: $f_{X,Y}(x,y) = 3x\,\mathbf 1_{0\leq y\leq x\leq 1}$

Let's look at the suggested answers.

A) $\def\d{\mathop{\rm d}} \displaystyle\int_0^x \int_y^1 3x^2 \d x\d y= x-\frac{x^4}{4}$

The bound variable for the inner integral appears in the bounds of the outer integral, and in the answer. That's clearly not right.

B) $\displaystyle\int_y^1\int_0^x3x^2 \d y\d x=-\frac{3}{4}(y^4-1)$

Likewise. Once you have evaluated the inner integral its bound variable (in this case, $y$) should no longer appear in the expression.

C) $\displaystyle\int_0^1 \int_0^1 3x^2 \d x \d y=\int_0^1 \int_0^1 3x^2 \d y\d x=1$

This has the opposite problem.   This is integrating outside the support; the interval here is $\{(x,y):x\in[0;1], y\in[0,1]\}$, which allows $x<y$.

D) $\displaystyle\int_0^1 \int_y^1 3x^2 \d x\d y=\frac{3}{4}$

This is correct.   This is integrating over the interval, $0\leq y\leq 1$ and $y\leq x\leq 1$.   The inner integral's bounds depend only on the variable for the outer integral, and the outer integrals bounds do not depend on either variable.

E) $\displaystyle\int_0^1 \int_0^x 3x^2 \d y\d x=\frac{3}{4}$

This is also correct.   This is integrating over $0\leq x\leq 1$ and $0\leq y\leq x$, which is the same interval as for $(D)$, just with a change of order of integration.

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The support of the joint distribution of $X$ and $Y$ is as shown below

enter image description here

the only right way of setting up the integral for the marginal density of $X$ is

$$f_X(x)=\int_0^x 3x \ dy =3x\left[y\right]_0^x=3x^2$$

if $0\le x\le 1$ and $0$ otherwise.

The expectation of $X$ is then

$$E[X]=3\int_0^1 x^3\ dx=\frac34.$$

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  • $\begingroup$ but the range of y is from 0 to x, shouldn't you derive the marginal probability like I did? $\endgroup$ – Amen Mar 10 '17 at 7:51
  • $\begingroup$ No wait, I edit. $\endgroup$ – zoli Mar 10 '17 at 7:54
  • $\begingroup$ I now realize that my last approach was a little wrong and it would result in the same answer as D so I deleted it. but as for your approach, shouldnt you compute the probability of all ys? $\endgroup$ – Amen Mar 10 '17 at 7:55
  • $\begingroup$ It's OK now, $\frac34$ is the solution. $\endgroup$ – zoli Mar 10 '17 at 7:59
  • $\begingroup$ Ok good, I agree that your solution is correct and I can understand it, what I don't understand is that why the D approach is correct and it would result in the same thing despite of ignoring the range for y? and isn't A the correct form of D? $\endgroup$ – Amen Mar 10 '17 at 8:04

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