Combined probability distribution

[Uncle_John]

Homework Statement

Let's have a box in shape of a square(viewed from the top) from the corner of which a smaller square was cut out.The side of a bigger square is 2a, side of the smaller square is a long.
We've got evenly distributed corn seeds all over the box,randomly selected seed is defined by coordinates $x,y \in [0,2a]$

Homework Equations

a.) Write down the combined probability distribution for $w(x,y)$
b.) Write down the projected probability distribution for $u(x)$(independent of $y$)
c.) calculate the correlation coefficient $r_{x,y}$

The Attempt at a Solution

a.) $1/3a^₂$
b.) $u(x)= 1/3a$ if $x \in [0,a]$
$u(x) = 2/3a$ if $x \in [a,2a]$
c.) since $r_{x,y} =\frac{\sigma_{x,y}}{\sigma_{x} \sigma_{y}}$, i calculated each variance seperately:
$\sigma_{x} = \int xu(x)dx$
$\sigma_{y} = \int yu(y)dx$
$\sigma_{x,y} = \int\int (x - \overline{x})(y - \overline{y})w(x,y)dxdy$

Is that right?

 

[LCKurtz]

Homework Statement

Let's have a box in shape of a square(viewed from the top) from the corner of which a smaller square was cut out.The side of a bigger square is 2a, side of the smaller square is a long.
We've got evenly distributed corn seeds all over the box,randomly selected seed is defined by coordinates $x,y \in [0,2a]$

Homework Equations

a.) Write down the combined probability distribution for $w(x,y)$
b.) Write down the projected probability distribution for $u(x)$(independent of $y$)
[/b]
a.) $1/3a^2$

What is in the denominator? You need parentheses.

I'm not sure what you mean by the "combined" probability distribution. If you mean the joint density function, you don't have it. It would be a function of x and y. This would be reflected in the domain when you write it carefully.

b.) $u(x)= 1/3a$ if $x \in [0,a]$
$u(x) = 2/3a$ if $x \in [a,2a]$

Yes, that is the marginal density of x (if you put correct parentheses in). And you get a symmetric formula for y.

 

[LCKurtz]

c.) since $r_{x,y} =\frac{\sigma_{x,y}}{\sigma_{x} \sigma_{y}}$, i calculated each variance seperately:
$\sigma_{x} = \int xu(x)dx$
$\sigma_{y} = \int yu(y)dx$
$\sigma_{x,y} = \int\int (x - \overline{x})(y - \overline{y})w(x,y)dxdy$

Is that right?

No. Your formulas for σ_x and σ_y are wrong. And you will need to be careful what limits you use on the last one.

 

[Uncle_John]

Yes, sorry, i meant joint probability distribution. So if i follow the formal definition:
$w_{x,y}(x,y) = w_{y|x}(x,y)w_{x}(x)$

Then:
$w_{x|y}(x,y) = \begin{cases} 1/a, \text{if } x \in [0,a) \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$

Also, $w_{x}$ is known:

$w_{x}(x) = \begin{cases} 1/(3a), \text{if} x \in [0,a) \\ 2/(3a), \text{if} x \in [a,2a] \end{cases}$

Is that better?

 

[LCKurtz]

Yes, sorry, i meant joint probability distribution. So if i follow the formal definition:
$w_{x,y}(x,y) = w_{y|x}(x,y)w_{x}(x)$

Then:
$w_{x|y}(x,y) = \begin{cases} 1/a, \text{if } x \in [0,a) \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$

Is that better?

But you need to be more careful here. The conditional density of x|y depends on both x and y. It matters whether y is in (0,a) or (a,2a).

 

[Uncle_John]

$w_{x|y}(x,y) = \begin{cases} 0, \text{if} x \in [0,a) \text{and} y \in [0,a] \\ 1/a, \text{if } x \in [0,a) \text{and} y \in [a,2a] \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$

? Better. But how should i write the final answer? $w_{x,y}$
What is wrong with $\sigma$?

 

[LCKurtz]

$w_{x|y}(x,y) = \begin{cases} 0, \text{if} x \in [0,a) \text{and} y \in [0,a] \\ 1/a, \text{if } x \in [0,a) \text{and} y \in [a,2a] \\ 1/(2a), \text{if } x\in [a,2a] \end{cases}$

? Better. But how should i write the final answer? $w_{x,y}$
What is wrong with $\sigma$?

Yes, that' s better for the conditional density. But perhaps I was a bit cryptic in my earlier comments about the joint density. Your formula of 1/(3a²) was correct but it gives the appearance of not depending on x or y. You need to indicate the proper (x,y) domain and you will have it without going this conditional density stuff.

The formulas you have written for the σ's look like formulas for the means instead of the standard deviations.

 

[Uncle_John]

Ok, so it would look something like :

$w_{x,y}(x,y) = \begin{cases} 0, \text{if } x \in [0,a] \text{and } y \in [0,a] \\ 1/(3a^₂), \text{otherwise} \end{cases}$


$\sigma_{x}^2 = \overline{x^2} - \overline{x}^₂$

and for $\sigma_{x,y} = \int \int (x - \overline{x})(y - \overline{y}) w_{x,y} dx dy$

i integrate over [0,a]x [a,2a]first, and then over [a,2a]x [0,2a], is that allright?

Can i post other probability problems in here or should i open a new thread?

 

[LCKurtz]

Ok, so it would look something like :

$w_{x,y}(x,y) = \begin{cases} 0, \text{if } x \in [0,a] \text{and } y \in [0,a] \\ 1/(3a^₂), \text{otherwise} \end{cases}$

I know you understand it, but the "otherwise" above doesn't include x or y greater than 2a or less than 0

$\sigma_{x}^2 = \overline{x^2} - \overline{x}^₂$

and for $\sigma_{x,y} = \int \int (x - \overline{x})(y - \overline{y}) w_{x,y} dx dy$

i integrate over [0,a]x [a,2a]first, and then over [a,2a]x [0,2a], is that allright?

Yes, that is correct.

Can i post other probability problems in here or should i open a new thread?

You should start a new thread. Others are more likely to join a new thread and it's general forum policy anyway.