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Showing 2 functions are Probability Denisty Function

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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.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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I can not follow your method.

The definite integral of zero is zero. You need to integrate the functions for the area where they are not identically zero.
You need to prove that the integral of f(x,y) for the (x,y) plane is 1. If you prove something, never start with the statement to be proved.

in case a), f(x,y) is zero everywhere except the domain D:{ 0≤x≤2, 0≤y≤2} . The definite integral of zero is zero, so you have to perform the integration on D:
[tex]\int_0^2\int_0^2{\frac{1}{4} xy dx dy}[/tex].
When integrating with respect to x, y can be considered constant so
[tex]\int_0^2\int_0^2{\frac{1}{4} xy dx dy}=\frac{1}{4}\int_0^2{y(\int_0^2{xdx})dy}=\frac{1}{4}\int_0^2{2ydy}=\frac{1}{4}4=1[/tex].
In case b, the integral is

[tex]\int_0^{\infty}\int_0^{\infty}{e^{-x-y}dxdy}[/tex]

The integrand can be decomposed into an x-dependent and an y-dependent factor. [tex]\int_0^{\infty}\int_0^{\infty}{e^{-x}e^{-y}dxdy}[/tex]

Again, when integrating with respect to x, the y-dependent part can be considered constant and pulled out from the integrand.
[tex]\int_0^{\infty}{e^{-y}(\int_0^{\infty}{e^{-x}dx})dy}[/tex]

What is the definite integral of e-x?

ehild
 
  • #3
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Oh I'm sorry.
I meant to have that as a want.

The definite integral of e[itex]^{-x}[/itex] over the interval [0,∞) is -e[itex]^{-∞}[/itex] + e[itex]^{0}[/itex] or 1, and the same thing for e[itex]^{-y}[/itex].
 

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