- #1

Ultramilk

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1. Homework Statement

Let functions f(x,y) with the following properties be defined as Probability Density Functions:

a) f(x,y) ≥ 0 for all (x,y)

b) ∫ℝ2 f dA = 1 (the ℝ2 is a subscript of the integral)

x and y are called independent random variables.

a) Show f(x,y) = { 1/4xy 0≤ x,y ≤ 2 ; 0 otherwise }

a) Show f(x,y) = { e^(−x−y) x,y≥ 0; 0 otherwise }

2. Homework Equations

3. The Attempt at a Solution

a+b) Proving the a) of the probability density function for both parts is the easy part. For both functions under otherwise f(x,y) = 0. For the first function 1/4xy, the minimum value under the boundary 0 ≤ x,y ≤ 2 is 0, and the exponential function e is a positive function.

Proving b is harder (I guess).

So what I did first is tried to tackle the problem when f(x,y) = 0.

∫ℝ2 f dA = 1

I'm not too particularly sure what this meant, so I'm assuming

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫f dy dx = 1

So, doing all that, I got:

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫0 dy dx = 1

Well I have to show the above.

But here's where I hit my first obstacle.

[itex]^{b}_{a}[/itex]∫0 dy is 0y |[itex]^{b}_{a}[/itex]

so 0b - 0a = 0

Fine. Moving on

[itex]^{d}_{c}[/itex]∫0 dx is 0x |[itex]^{d}_{c}[/itex]

again 0d - 0c = 0.

I have a feeling that I did something wrong. The wording of the question implies that both of these piecewise are probability density functions. Is there a property that says

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫0 dy dx = 1?

_____________________________________________

Okay now the functions.

[itex]^{2}_{0}[/itex]∫[itex]^{2}_{0}[/itex]∫1/4xy dy dx = 1

∫[itex]^{2}_{0}[/itex]∫1/4xy dy = 1/4 x/2 y^2 |[itex]^{2}_{0}[/itex]

=

∫[itex]^{2}_{0}[/itex]∫x/2 dx = (x^2)/2 |[itex]^{2}_{0}[/itex]

= 1.

This part, I just want confirmation of my work.

___________________________________________________________

Now the last one:

[itex]^{∞}_{0}[/itex]∫[itex]^{∞}_{0}[/itex]∫e^(-x-y) dy dx = 1

The moment I evaluate:

∫[itex]^{∞}_{0}[/itex]∫e^(-x-y) dy

it becomes

-e^(-x-y)|[itex]^{∞}_{0}[/itex]

-e^(-x-∞) + e^(-x) < --- (unsure if this is right at all!)

[itex]^{∞}_{0}[/itex]∫-e^(-x-∞) + e^(-x) dx

e^(-x-∞) - e^(-x) |[itex]^{∞}_{0}[/itex]

e^(-∞-∞) - e^(-∞) -(e^(-∞) - e^(0)) ?= 1

but idk if e^(-2∞) - 2e(-∞) + 1 = 1

I felt I did something wrong again.

EDIT: Oh stupid me (I did another problem before I realized this)

e^(-2∞) - 2e(-∞) = 0?

because you can rewrite it as 1/e^(2∞) - 2/e^(∞). Both values are incredibly small, so it's 0. So 1=1. Yes?

Thanks a lot for the help.

Sorry about the use of the symbols, I'm not super adept at it.

Let functions f(x,y) with the following properties be defined as Probability Density Functions:

a) f(x,y) ≥ 0 for all (x,y)

b) ∫ℝ2 f dA = 1 (the ℝ2 is a subscript of the integral)

x and y are called independent random variables.

a) Show f(x,y) = { 1/4xy 0≤ x,y ≤ 2 ; 0 otherwise }

a) Show f(x,y) = { e^(−x−y) x,y≥ 0; 0 otherwise }

2. Homework Equations

3. The Attempt at a Solution

a+b) Proving the a) of the probability density function for both parts is the easy part. For both functions under otherwise f(x,y) = 0. For the first function 1/4xy, the minimum value under the boundary 0 ≤ x,y ≤ 2 is 0, and the exponential function e is a positive function.

Proving b is harder (I guess).

So what I did first is tried to tackle the problem when f(x,y) = 0.

∫ℝ2 f dA = 1

I'm not too particularly sure what this meant, so I'm assuming

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫f dy dx = 1

So, doing all that, I got:

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫0 dy dx = 1

Well I have to show the above.

But here's where I hit my first obstacle.

[itex]^{b}_{a}[/itex]∫0 dy is 0y |[itex]^{b}_{a}[/itex]

so 0b - 0a = 0

Fine. Moving on

[itex]^{d}_{c}[/itex]∫0 dx is 0x |[itex]^{d}_{c}[/itex]

again 0d - 0c = 0.

I have a feeling that I did something wrong. The wording of the question implies that both of these piecewise are probability density functions. Is there a property that says

[itex]^{d}_{c}[/itex]∫[itex]^{b}_{a}[/itex]∫0 dy dx = 1?

_____________________________________________

Okay now the functions.

[itex]^{2}_{0}[/itex]∫[itex]^{2}_{0}[/itex]∫1/4xy dy dx = 1

∫[itex]^{2}_{0}[/itex]∫1/4xy dy = 1/4 x/2 y^2 |[itex]^{2}_{0}[/itex]

=

∫[itex]^{2}_{0}[/itex]∫x/2 dx = (x^2)/2 |[itex]^{2}_{0}[/itex]

= 1.

This part, I just want confirmation of my work.

___________________________________________________________

Now the last one:

[itex]^{∞}_{0}[/itex]∫[itex]^{∞}_{0}[/itex]∫e^(-x-y) dy dx = 1

The moment I evaluate:

∫[itex]^{∞}_{0}[/itex]∫e^(-x-y) dy

it becomes

-e^(-x-y)|[itex]^{∞}_{0}[/itex]

-e^(-x-∞) + e^(-x) < --- (unsure if this is right at all!)

[itex]^{∞}_{0}[/itex]∫-e^(-x-∞) + e^(-x) dx

e^(-x-∞) - e^(-x) |[itex]^{∞}_{0}[/itex]

e^(-∞-∞) - e^(-∞) -(e^(-∞) - e^(0)) ?= 1

but idk if e^(-2∞) - 2e(-∞) + 1 = 1

I felt I did something wrong again.

EDIT: Oh stupid me (I did another problem before I realized this)

e^(-2∞) - 2e(-∞) = 0?

because you can rewrite it as 1/e^(2∞) - 2/e^(∞). Both values are incredibly small, so it's 0. So 1=1. Yes?

Thanks a lot for the help.

Sorry about the use of the symbols, I'm not super adept at it.

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