I am a newbie in the field of probability theory and am trying to understand the mechanics for computing joint probability density function and everything related to this subject. As such, I was trying to solve the following problem: $$f(x,y) = \begin{cases} cxy^2(1-x)(1-y^2), & 0< x\leq 1,0< y\leq 1 \\ 0, & \text{ otherwise} \end{cases}$$

From what I have understood, the answer is computed by:

$$\int\limits_{0}^{1}\int\limits_{0}^{1}cxy^2(1-x)(1-y^2)\, dxdy = 1 $$ First, I have noticed that on some web resources, the upper limit y is given as x instead of 1 and I do not quite understand why it is so and would appreciate if someone could explain this to me. Second, I would like to understand the mechanics of calculating double integrals such as this, after exhaustive research on the web, all I have noticed, is that there is no uniform way of doing it, and now please correct me if I am wrong but the way I have proceeded, was by initially treating x as a constant and thus writing it as: $$c[\int\limits_{0}^{1}x(1-x)dx\int\limits_{0}^{1}y^2(1-y^2)\, dy]=1 $$ Then I integrate y followed by x which will lead me to compute c=45

Are there any flaws with this method?

Tank you!


$$\int\limits_{0}^{1}\int\limits_{0}^{1}cxy^2(1-x)(1-y^2)\, dxdy =1$$ Then find c from this.

You are correct in your value of c = 45.

The only time the limit of integral becomes x is when you have the intervals for x and y to be $ 0<y<x$ and $0<x<1$.

As far as this problem goes, you will have limits running from 0 to 1 for both x and y.

While figuring out the limits, you have to look at the intervals for which f(x,y) is defined for both x and y

Suppose if the function was defined as this

$$f(x,y) = \begin{cases} cxy^2(1-x)(1-y^2), & 0< x\leq 1,0< y\leq x \\ 0, & \text{ otherwise} \end{cases}$$


$$\int\limits_{0}^{1}\int\limits_{0}^{x}cxy^2(1-x)(1-y^2)\, dydx =1$$

That will be the one and since the of y is dependent on x, you have to integrate y first keeping x constant and then, substitute the value of x in the integrated y and now collectively integrate for x. That should be the way.

Good luck


  • $\begingroup$ I am sorry, I omitted to equal the entire thing to 1, I will edit it right away. However, would you mind, elaborating on the last part of your answer, namely when you say: "While figuring out the limits, you have to look at the intervals for which f(x,y) is defined for both x and y" $\endgroup$ – O.A. Feb 14 '15 at 21:20

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