Given marginal pdfs of X and Y, find pdf of Z=X-Y

In summary, Ray tried to solve a problem where the new variable was to be found by marginalizing over the marginals, but wasn't able to do so. He suggests looking at the sum of two independent distributions as a way to get to the new variable. Arcana provides a solution for adding or subtracting independent distributions. Finally, the probability density function of U is given.
  • #1
ArcanaNoir
779
4

Homework Statement



The probability density function of the random variables X and Y are given by:

[tex] f_1(x)= \begin{cases} 2 & -\frac{1}{4}\le x\le \frac{1}{4} \\ 0 & \text{elsewhere} \end{cases} [/tex]
and
[tex] f_2(y) \begin{cases} \frac{1}{2} & 0\le y \le 2 \\ 0 & \text{elsewhere} \end{cases} [/tex]

respectively.

a) Find the probability density function of the random variable Z=X-Y .
b) What is the probability that Z will assume a value greater than zero?

Homework Equations



Not sure yet.

The Attempt at a Solution



There isn't an example like this in my book. I'm not sure how to go from marginals to the new variable thing, which I couldn't solve in an ordinary manner anyway! Sad sad sad. Am I supposed to make the marginals into a regular f(x,y), or is there some direct way to get to the Z?
 
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  • #2
I assume your book tells you how to compute the distribution of a sum of random variables such as W=X+Y.

One way to look at this is to invent a new random variable U=-Y. (Use Z=X-Y=X+(-Y)=X+U.) What does the distribution of this variable U look like? of X+U?
 
  • #3
ArcanaNoir said:

Homework Statement



The probability density function of the random variables X and Y are given by:

[tex] f_1(x)= \begin{cases} 2 & -\frac{1}{4}\le x\le \frac{1}{4} \\ 0 & \text{elsewhere} \end{cases} [/tex]
and
[tex] f_2(y) \begin{cases} \frac{1}{2} & 0\le y \le 2 \\ 0 & \text{elsewhere} \end{cases} [/tex]

respectively.

a) Find the probability density function of the random variable Z=X-Y .
b) What is the probability that Z will assume a value greater than zero?


Homework Equations



Not sure yet.

The Attempt at a Solution



There isn't an example like this in my book. I'm not sure how to go from marginals to the new variable thing, which I couldn't solve in an ordinary manner anyway! Sad sad sad. Am I supposed to make the marginals into a regular f(x,y), or is there some direct way to get to the Z?

Unless you are given more information you cannot do the question:you need to know something about the joint distribution of the pair (X,Y). In particular, are X and Y independent? If they *are* independent, just let Y1 = -Y and look at X+Y1. The distribution of Y1 is easy to get, and surely the distribution of X+Y1 must be obtainable from material in your textbook or notes.

RGV
 
  • #4
Ray Vickson said:
Unless you are given more information you cannot do the question:you need to know something about the joint distribution of the pair (X,Y). In particular, are X and Y independent? If they *are* independent, just let Y1 = -Y and look at X+Y1. The distribution of Y1 is easy to get, and surely the distribution of X+Y1 must be obtainable from material in your textbook or notes.

RGV

What I typed is all I have.
 
  • #5
So assume they are independent. As both Ray and I noted, your text or notes must have something to say about the sum of two independent random variables.
 
  • #6
Hmm, it looks like if they are independent then [itex] f(x,y)=f_1(x)f_2(y) [/itex]
From there, it's like any other random variable problem. Thanks for the suggestion. :)
 
  • #7
Hi Arcana! :smile:

For adding or subtracting independent distributions, we have the convolution rule for distributions.

Suppose X and Y are independent probability distributions with probability density functions fX(x) and fY(y), and cumulative probability function FX(x) and FY(y).

If U=X+Y, then
[tex]P(U \le u)
= P(X + Y \le u)
= \int_{-\infty}^{\infty} f_X(x) P(x+Y \le u) \textrm{ d}x
= \int_{-\infty}^{\infty} f_X(x) P(Y \le u - x) \textrm{ d}x
[/tex]
so
[tex]P(U \le u)
= \int_{-\infty}^{\infty} f_X(x) F_Y(u-x) \textrm{ d}x
[/tex]

And if you want to know the probability density of U, we have:
[tex]f_U(u)= {d \over du}F_U(u) = {d \over du}P(U \le u)[/tex]
 
  • #8
great, thanks!
 

FAQ: Given marginal pdfs of X and Y, find pdf of Z=X-Y

1. What is a marginal pdf?

A marginal pdf, or probability density function, is a mathematical function that describes the probability of a continuous random variable taking on a specific value within a given range.

2. How is the pdf of Z=X-Y calculated?

The pdf of Z can be calculated by taking the convolution of the marginal pdfs of X and Y. This involves integrating the product of the two pdfs over all possible values of Z.

3. Can the pdf of Z be negative?

No, the pdf of Z cannot be negative as it represents the probability of Z taking on a specific value. The pdf must always be non-negative.

4. What is the relationship between the pdf of Z and the joint pdf of X and Y?

The joint pdf of X and Y describes the probability of both X and Y taking on specific values simultaneously. The pdf of Z is derived from the joint pdf and represents the probability of the difference between X and Y.

5. Are there any assumptions or limitations when using the marginal pdfs to find the pdf of Z?

Yes, the marginal pdfs should be independent and the variables should have a continuous distribution. Additionally, the marginal pdfs should be defined over the same range of values. Failure to meet these assumptions may result in incorrect calculations.

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