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I've been working on this assignment:

Let (X, Y) be a continuous bivariate r.v. with joint pdf

$$ f_{XY}(x,y) = e^{-(x+y)} \: x>0, y>0$$ $$ f_{XY}(x,y) = 0 \: otherwise$$

i calculated the joint moment generating function, it look like this:$$M_{xy}(t_1,t_2) = \frac{1}{(1-t_1)(1-t_2)} $$ but now it's asking me to calculate the joint momentum for $m_{1,0},m_{0,1},m_{1,1}$ Acording to my textbook the answer for this 3 is: $m_{1,0}=1,m_{0,1}=1,m_{1,1}=1$

but i don't understand how to calculate this, i can't understand the formula: $$ m_{k,n} = E(X^k Y^n) = M_{XY}^{(kn)} (0,0) $$

can someone explain to me how to use this formula?, (maybe for different momentums )

Eduardo Gutierrez
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1 Answers1

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$m_{k,n}$

Symbols $k,n$ are indices here to denote different moments. If you need to calculate moment $m_{2,3}$, then $k=2$ and $n=3$

$E(X^kY^n)$

Upper indices $k,n$ here are powers. The formula reads expectation of a product $Z=X^kY^n$.

$M_{XY}^{(kn)}$

It can be a typo, or an unfortunate notation. It should better be written as $M_{XY}^{(k),(n)}$, where it means that this is $k-th$ derivative on first argument, and $n-th$ derivative on the second argument: $$ M_{XY}^{(2),(3)}(t_1,t_2) = \frac{\partial^5}{\partial^2 t_1\partial^3 t_2} M_{XY}(t_1,t_2). $$ When you have a specific point in parentheses $(0,0)$, it means you should take the value of this derivative at point $t_1=t_2=0$.

Vasily Mitch
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