## Expected Value and Multiple Integrals

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MuthuVeerappanR
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### Expected Value and Multiple Integrals

G'Morning all, I have a couple of questions regarding geometric probability.

Consider a point selected uniformly at random from a right angle triangle bounded by the lines $x=0$, $x=a$, and $y=mx$. Say I want to calculate the expected value of some function $f(x,y)$. Is it right to do it the following way?

$\mathbb{E}(f(x, y))=\displaystyle\frac{2}{ab}\int\limits_{0}^a\int\limits_{0}^{m x}f(x,y)\,dy\,dx$

Had it been a rectangle I would have no doubts about the integral. But in a right-angled triangle, this seems like doing the following.

Pick a value $x$ uniformly between $0$ and $a$. Now pick $y$ from $0$ to $mx$.

The second statement is where I have my doubts. Is it right to take that $y$ so selected will be uniform?

In cases like this, how do we check whether that particular value is uniformly distributed?

Say, If have $\mathbb{E}(f(x, y))$. Then is there a easier way to find any of $\mathbb{E}(f(a - x, y))$, $\mathbb{E}(f(x, m a - y))$, or $\mathbb{E}(f(a - x, m a - y))$??

Thanks for the help.

I'm not asking to this to solve any PE question. Also, I haven't solved any PE questions using these. So I don't think answering them wont be a spoiler.

It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment.

jaap
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### Re: Expected Value and Multiple Integrals

I'm no expert on this, but I think it is correct. Basically, you have a sample space consisting of the points in the 2-dimensional plane $\Bbb R^2$, and a uniform Probability density function which has the value $\frac{2}{ab}$ for every point inside the triangle $T$, and $0$ outside of it:

$$pdf(P) = \begin{cases} \frac{2}{ab}, & \text{if P \in T} \\ 0, & \text{if P \notin T} \end{cases}$$

It is a valid pdf because its integral equals $1$:

$$\int\limits_{P \in \Bbb R^2} pdf(P) dP = \int\limits_{P \in T} \frac{2}{ab} dP = \frac{2}{ab} \int\limits_{P \in T} 1 dP = \frac{2}{ab} \frac{ab}{2} = 1$$

The expected value of some function $f(P)$ is then the integral of the pdf times the function:

$$\int\limits_{P \in \Bbb R^2} pdf(P) f(P) dP = \int\limits_{P \in T} \frac{2}{ab}f(P) dP$$

It does not matter how you do the integration over the triangle. Each part of the triangle contributes proportional to its area.

MuthuVeerappanR
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### Re: Expected Value and Multiple Integrals

Thanks for the explanation jaap. After thinking about it today, I think I get that the points will be uniformly distributed. So kind of got clarity on the first question.

Again, anyone.. Any thoughts on the Second part of the question?? I don't think there is one but it surprises me that the analogy in single integral case is so simple to answer.

It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment.

jaap
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### Re: Expected Value and Multiple Integrals

MuthuVeerappanR wrote:
Fri Jun 09, 2017 4:14 am
Say, If have $\mathbb{E}(f(x, y))$. Then is there a easier way to find any of $\mathbb{E}(f(a - x, y))$, $\mathbb{E}(f(x, m a - y))$, or $\mathbb{E}(f(a - x, m a - y))$??
No.
$\mathbb{E}(f(x, y)|(x,y) \in T)$ is the expected value that $f$ has within the triangle $T$, which is bounded by the lines $y=0$, $x=a$, and $y=mx$.

$\mathbb{E}(f(a-x, y)|(x,y) \in T) = \mathbb{E}(f(x, y)|(x,y) \in T_2)$ is the expected value that $f$ has within a triangle $T_2$, which is bounded by the lines $y=0$, $x=0$, and $y=m(a-x)$.

$\mathbb{E}(f(x, ma - y)|(x,y) \in T) = \mathbb{E}(f(x, y)|(x,y) \in T_3)$ is the expected value that $f$ has within a triangle $T_3$, which is bounded by the lines $y=ma$, $x=0$, and $y=ma-mx$.

$\mathbb{E}(f(a-x, ma - y)|(x,y) \in T) = \mathbb{E}(f(x, y)|(x,y) \in T_4)$ is the expected value that $f$ has within a triangle $T_4$, which is bounded by the lines $y=ma$, $x=a$, and $y=mx$.

These are different regions of the plane, so unless your function $f$ has some properties that relate the values over these triangles, there is not a lot you can do. However, $T \cup T_4 = T2 \cup T_3$, so once you know three of these you can deduce the fourth.

MuthuVeerappanR
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### Re: Expected Value and Multiple Integrals

That's nice jaap... Thanks a lot..

It is not knowledge, but the act of learning, not possession but the act of getting there, which grants the greatest enjoyment.