0
$\begingroup$

I was hoping somebody could explain to me how you go about determining the domain for the marginal distribution, the marginal expectations, and the marginal variances. Here is the problem:

Let $f(x,y)= \left\lbrace \begin{matrix} 2 & 0\le y\le x \le 1 \\ 0 & otherwise \end{matrix} \right. $

Find $μ_x, \sigma^2_x,$ and $f(x)$.

I don't need somebody to solve this for me, as I already have the solution. However, I do not really understand how the domains of these three functions work. Of a similar vein, I do not understand which limits of integration to use at each step throughout the problem. Specifically, I'm trying to figure out:

  1. When solving the integral for $f(x)=\int f(x,y)dy$, it is intuitive to me that you would integrate between $0$ and $x$. However, I do not understand why the domain for $f(x)$ is $0\le x \le 1$.
  2. After you solve for the expected value of $x$, why is its domain $0\le x \le 1$?
  3. What would the domain be for the variance of $x$? Is there an intuitive way to understand the domain for a marginal variance where the original function's domain for $x$ depends on $y$ (and vice versa), such as this one does?

I would greatly appreciate anybody who could clear this up for me. I have been unable to find a good explanation for these questions in my textbook. Thanks in advance!

$\endgroup$
1
$\begingroup$

Consider your pdf: $~f_{X,Y}(x,y)= 2\,\mathbf 1_{0\leq x\leq y\leq 1}$.   Where $\mathbf 1_A$ is an indicator, having value of $1$ when the condition $A$ is true, and $0$ otherwise.

Notice that the $f_{X,Y}$'s support of $\{(x,y): 0\leq x\leq y\leq 1\}$ is the triangle $\triangle(0,0)(0,1)(1,1)$.

Now to find the marginal for $X$ we must 'integrate out' the variable $Y$ for any given value for $X$.   Thus the bounds of this integral are $x\leq y\leq 1$.   Having 'integrated out' $y$ this leaves us with a support for $f_X$ of $\{x:0\leq x\leq 1\}$.   This is the projection of the triangle onto the x-axis.

$$f_X(x)~{=~\int_\Bbb R 2\;\mathbf 1_{0\leq x\leq y\leq 1 } \operatorname d y\\ =~2\;\mathbf 1_{0\leq x\leq 1}\int_x^1\operatorname d y\\=~2\,(1-x)\;\mathbf 1_{0\leq x\leq 1}}$$

Now we obtain the expected value for $X$ and $X^2$ by integrating over the support for its marginal pdf, via $$\mathbb E(X^n)~{=~\int_\Bbb R x^n\cdot f_X(x)\operatorname d x\\ =~ 2 \int_0^1 x^n(1-x)\operatorname d x \\ =~ 2(n+1)^{-1}(n+2)^{-1}}\quad\forall n\in\Bbb N$$

Thus $\mu_X=\tfrac 1 3~,~ \sigma_X^2=\tfrac 1{18}$

$\endgroup$
  • $\begingroup$ Unfortunately, I have never encountered the approach that you are using in your explanation, so I am not sure this makes it any easier for me to understand. I appreciate your time & help regardless! $\endgroup$ – Hawleyluyah Jun 11 '17 at 2:58
  • $\begingroup$ Actually, this is beginning to make a bit of sense to me. So after you integrate out y, the domain of f(x) is the same except that you ignore the y component? Furthermore, the domain of the marginal expected value and variance (of x) would be the same as the domain of f(x)? $\endgroup$ – Hawleyluyah Jun 11 '17 at 3:05
  • $\begingroup$ @Hawleyluyah Indeed, after 'integrating out' $y$, then $y$ in no longer a component of the support. However, the mean and variance themselves do no have a support; they are constant values. $\endgroup$ – Graham Kemp Jun 11 '17 at 3:53

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.