Transformation Of Probability Density Functions

In summary, the problem involves finding the pdf for Z, where X and Y are random variables with pdfs f_X(x)=2(1-x) and f_Y(y)=2(1-y) respectively, defined on [0,1]. The solution involves splitting the problem into different cases and integrating over the area under the curve y=z-x. This takes into account the fact that X and Y may not be independent.
  • #1
caffeine

Homework Statement


Let X and Y be random variables. The pdfs are [itex]f_X(x)=2(1-x)[/itex] and [itex]f_Y(y) = 2(1-y)[/itex]. Both distributions are defined on [0,1].

Let Z = X + Y. Find the pdf for Z, [itex]f_Z(z)[/itex].


Homework Equations


I'm using ideas, not equations.


The Attempt at a Solution


I'm dying of curiosity about where I'm going wrong. I'm so sure of each step, but my answer can't be correct because [itex]\int_0^2 f_Z(z)\,dz[/itex] is zero! Here's my logic.

Consider the cdf (cumulative distribution function) for Z:

[tex]
F_Z(z) = P(Z\le z) = P(X+Y \le z)
[/tex]

Here, [itex]F_Z(z)[/itex] is the volume above the triangle shown in the image I attached to this message (in case something happens to the attachment, it's the triangle in quadrant 1 bounded by x=0, y=0 and x+y=z.)

The volume above the shaded region represents [itex]F_Z(z)[/itex].

[tex]
F_Z(z) = \int_{x=0}^{x=z} 2(1-x)\int_{y=0}^{y=z-x} 2(1-y)\,dy\,dx
[/tex]

Performing the integrals gives [itex]F_Z(z) = \frac{1}{6}z^4 - \frac{4}{3}z^3 + 2z^2[/itex]. Then taking the derivative of the cdf gives the pdf:

[tex]
f_Z(z) = \partial_z F_Z(z) = \frac{2}{3}z^3 - 4z^2 +4z
[/tex]

Unfortunately, this can't be right because the integral of this function over [0,2] gives zero.

I also would've expected that the maximum of [itex]f_Z(z)[/itex] would be at z=0 since individually, X and Y are most likely to be zero.

I checked my algebra and calculus with Mathematica; it looked fine. I think there's a conceptual problem. There must be something I don't understand or some point I'm not clear about.

What did I do wrong?
 

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  • #2
You have to split the problem into different cases.

i) z<0
ii) z<1
iii) 1<z<2
iv) z>2

In each case the area of integration is different. But for each case, your model for the integral is

[tex]F_Z(z)=\int\int_{\{(x,y)\in [0,1]\times[0,1]:\ y\leq z-x, \}}f_{XY}(x,y)dxdy[/tex]

So you're integrating over the area that's the intersection of the square [0,1] x [0,1] with the area under the curve y=z-x.(Btw, you never said that X and Y are independant but I assume they are?)
 
Last edited:
  • #3
Ahhh... yes. I *completely* understand.

We're integrating that portion of the unit square (in QI with lower left corner at origin) that lies underneath x+y=1.

So when 0<z<1, there's only one function that represents the "top": the line x+y=z.

And when 1<z<2, there's two functions: y=1 and x+y=z, and therefore we need to break the region of integration into two portions.

Got it. Thanks.

And, yes. :-)
 

What is the transformation of probability density functions?

The transformation of probability density functions refers to the process of changing the probability distribution of a random variable by applying a mathematical function to it. This results in a new probability density function that describes the distribution of the transformed variable.

Why is it important to understand transformation of probability density functions?

Understanding transformation of probability density functions is important because it allows us to model and analyze complex systems and phenomena in a more accurate and efficient manner. It also enables us to make predictions and draw conclusions based on the transformed data.

What are the most common transformations used for probability density functions?

The most commonly used transformations for probability density functions include logarithmic, exponential, power, and linear transformations. Each of these transformations has its own unique properties and is suitable for different types of data.

Can any function be used to transform a probability density function?

No, not all functions can be used to transform a probability density function. The function must be one-to-one and differentiable in order to ensure that the transformed variable has a valid probability distribution. Additionally, the function should also preserve the order of the data points.

How is the transformation of probability density functions related to the Central Limit Theorem?

The transformation of probability density functions plays a crucial role in the Central Limit Theorem. This theorem states that when independent random variables are added together, their sum will tend towards a normal distribution. The transformation of probability density functions allows us to transform any distribution into a normal distribution, making it a powerful tool in statistical analysis.

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