I wanted to ask if the order of integration matters in double integration when we are calculating expectations or joint probability etc. For example: fXY (x, y) = 8xy, 0 ≤ x ≤ 1, 0 ≤ y ≤ x and 0 otherwise

I had to compute the E(Y) and to do that I first found the marginal distribution of y which came out as 4y and then I found its expectation which was (4/3)x^3 I also tried finding the expectation directly using the distribution function integrating first w.r.t y and then w.r.t x and the answer came out 8/15

Which answer is correct? And why are both methods giving me different answers? Also, had I integrated w.r.t x first and then y, I'd have gotten the the same answer I did when I found the marginal pdf and then found the expectation.

So, does the order of integration matter?


The map $(x,y)\to 8xy$ is continuous, and the region $$ \mathcal D:=\{(x,y)\in\mathbb R^2 : 0\leqslant x\leqslant 1, 0\leqslant y\leqslant x \} $$ is both closed and bounded, and hence compact. It follows that $f_{X,Y}$ is bounded, and so $$ \int_{\mathcal D}|f_{X,Y}|\ \mathsf d(x\times y)<\infty. $$ Hence by Fubini's theorem, the iterated integrals $$ \int_{\mathbb R}\int_{\mathbb R}f_{X,Y}(x,y)\ \mathsf dx\ \mathsf dy $$ and $$ \int_{\mathbb R}\int_{\mathbb R}f_{X,Y}(x,y)\ \mathsf dy\ \mathsf dx $$ exist, and are equal. So the order of integration does not matter. To compute the marginal density of $Y$, we integrate $f_{X,Y}$ over all values of $x$: $$ f_Y(y) = \int_y^1 8xy\ \mathsf dx = 4y(1-y^2). $$ We compute the expectation of $Y$ by integrating $yf_Y(y)$: $$ \mathbb E[Y] = \int_0^1 4y^2(1-y^2)\ \mathsf dy = \frac8{15}. $$ Your error was in the bounds of the integral for computing $f_Y$ - the lower bound should be $y$ since since $x\leqslant y$ in the region $\mathcal D$.


You have.$$f_{X,Y} (x, y) = \begin{cases}8xy&,& 0 ≤ x ≤ 1, 0 ≤ y ≤ x \\ 0 && \textsf{otherwise}\end{cases}$$

The order of integration does not matter (see Fubini' and Tonelli's theorems). What matters is using the correct integration domains.

$$\begin{split}\mathsf E(Y) &= \iint_{0\leqslant y\leqslant x\leqslant 1} y\cdot(8xy)~\mathsf d(x,y)\\&= \int_0^1\int_0^x 8xy^2~\mathsf d y~\mathsf d x \\&= \int_0^1\int_y^1 8xy^2~\mathsf d x~\mathsf d y \\&= \tfrac 8{15}\end{split}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.