I'm given the following joint pdf:

$$ f(A,B)= \begin{cases} c(a^2+b) & \text{ for } 0 \leq b \leq 1-a^2\\ 0 & \text{ else} \end{cases} $$

I'm supposed to find c but I'm not sure how to find the limits of integration. I thought that they were 0 and 1 for db and -1 and 1 for da, giving me 5/3 when I integrate, but that isn't correct. Could anyone help me understand what's going wrong?

Overall, I just having difficulty understand the limits of integration when the other variable is in the equation, for example Pr(B>X+1) or Pr(B=X^1/2).


If $|a| > 1$, then $1-a^2 < 0$, and so, there are no values of $b$ such that $0 \le b \le 1-a^2$.

If $|a| \le 1$, then $1-a^2 \ge 0$, and so, there are values of $b$ such that $0 \le b \le 1-a^2$.

Hence, the bounds are $-1 \le a \le 1$ and $0 \le b \le 1-a^2$.

Notice that the upper bound for $b$ isn't simply $1$, it actually depends on $a$. This is because the region you are integrating over isn't a simple rectangle, but rather a curved region.

So, you need to find the value of $c$ such that $\displaystyle\int_{-1}^{1}\int_{0}^{1-a^2}c(a^2+b)\,db\,da = 1$.

  • $\begingroup$ The bounds still happen to be a bit unclear for me. So, what if b<=a+.5 instead, how would the bounds change? $\endgroup$ – nodel Jul 11 '15 at 9:07

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