Math Problem Statement

Determine the probability mass function (PMF) for the random variable Y based on the given probability mass function f(x) and transformations of Y.

Solution

From the given image, it appears to be a probability problem involving a probability mass function f(x)f(x) and transformations of a random variable YY. I will solve it step-by-step. Let’s address the task:

Given:

  1. ( f(x) = \begin{cases} p(1-p)^{x-1}, & x = 1, 2, 3, \dots \ 0, & \text{otherwise} \end{cases} )

  2. Different transformations for YY based on XX:

    • (a) Y=1x+1Y = \frac{1}{x+1}
    • (b) Y=x2Y = x^2
    • (c) ( Y = \begin{cases} (2 - |x - 3|), & x \leq 5 \ (x - 5), & x > 5 \end{cases} )

To Determine:

The probability mass function (PMF) of YY.

Let’s compute these:


(a) Y=1x+1Y = \frac{1}{x+1}

Here, YY is a function of XX. To derive the PMF of YY, we find the relationship between XX and YY. Specifically: Y=1x+1    x=1Y1Y = \frac{1}{x+1} \implies x = \frac{1}{Y} - 1 The PMF of YY will depend on substituting xx into f(x)f(x): P(Y=y)=f(x), where x=1y1P(Y = y) = f(x), \text{ where } x = \frac{1}{y} - 1 Thus: P(Y=y)=p(1p)1y2, for y=12,13,P(Y = y) = p(1-p)^{\frac{1}{y} - 2}, \text{ for } y = \frac{1}{2}, \frac{1}{3}, \dots


(b) Y=x2Y = x^2

Here, YY is a quadratic function of XX: Y=x2Y = x^2 Possible values of YY are 1,4,9,1, 4, 9, \dots (squares of positive integers). For each yy: P(Y=y)=P(X=y)P(Y = y) = P(X = \sqrt{y}) If y\sqrt{y} is an integer, then: P(Y=y)=f(y)=p(1p)y1P(Y = y) = f(\sqrt{y}) = p(1-p)^{\sqrt{y} - 1} Otherwise, P(Y=y)=0P(Y = y) = 0.


**(c) ( Y =

 \begin{cases} 
 (2 - |x - 3|), & x \leq 5 \\
 (x - 5), & x > 5 
 \end{cases} \)**

For x5x \leq 5: Y=2x3Y = 2 - |x - 3| Possible values of YY depend on the range of xx. Compute YY for x=1,2,3,4,5x = 1, 2, 3, 4, 5:

  • x=1x = 1: Y=0Y = 0
  • x=2x = 2: Y=1Y = 1
  • x=3x = 3: Y=2Y = 2
  • x=4x = 4: Y=1Y = 1
  • x=5x = 5: Y=0Y = 0

Thus, the PMF for YY for x5x \leq 5: P(Y=0)=f(1)+f(5),P(Y=1)=f(2)+f(4),P(Y=2)=f(3)P(Y = 0) = f(1) + f(5), \quad P(Y = 1) = f(2) + f(4), \quad P(Y = 2) = f(3)

For x>5x > 5: Y=x5Y = x - 5 Here, YY takes values 1,2,3,1, 2, 3, \dots (for x=6,7,8,x = 6, 7, 8, \dots): P(Y=y)=f(y+5)P(Y = y) = f(y + 5)


Let me know if you want further details, have any questions, or need these results expanded further.

Related Questions:

  1. What is the cumulative distribution function (CDF) of YY for each case?
  2. How does the PMF change if f(x)f(x) is redefined for a finite xx range?
  3. What is the expected value E[Y]E[Y] for each transformation?
  4. How would you compute the variance of YY in case (c)?
  5. How do these results generalize for other transformations of XX?

Tip:

Always carefully identify the relationship between the original random variable XX and the transformed random variable YY when working with PMFs.

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Math Problem Analysis

Mathematical Concepts

Probability mass function (PMF)
Geometric distribution
Transformation of random variables

Formulas

PMF of geometric distribution: f(x) = p(1-p)^(x-1), for x = 1, 2, 3, ...
Transformation of variables Y = g(X)

Theorems

Geometric distribution theorem
Transformation rule for discrete random variables

Suitable Grade Level

Undergraduate - Probability and Statistics