Math Problem Statement

the generating function of a discrete random variable is GX (t) = 1 − √1 − t, find the probabilities for P(X = k) for k = 1, 2, . . .. Explain why P(X = 0) = 0.

Solution

Given the generating function GX(t)=11tG_X(t) = 1 - \sqrt{1 - t}, we are tasked with finding the probabilities P(X=k)P(X = k) for k=1,2,k = 1, 2, \dots, as well as explaining why P(X=0)=0P(X = 0) = 0.

Step 1: Generating Function and Definition

The probability generating function GX(t)G_X(t) is defined as: GX(t)=k=0P(X=k)tkG_X(t) = \sum_{k=0}^{\infty} P(X = k) t^k This means that the coefficient of tkt^k in the series expansion of GX(t)G_X(t) gives the probability P(X=k)P(X = k).

Step 2: Expand the Generating Function

We start by expanding GX(t)=11tG_X(t) = 1 - \sqrt{1 - t} as a series in terms of tt.

First, recall that the square root function 1t\sqrt{1 - t} can be expanded using the binomial series (valid for t<1|t| < 1): 1t=n=0(12n)(t)n\sqrt{1 - t} = \sum_{n=0}^{\infty} \binom{\frac{1}{2}}{n} (-t)^n The first few terms of this expansion are: 1t=112t18t2116t3\sqrt{1 - t} = 1 - \frac{1}{2}t - \frac{1}{8}t^2 - \frac{1}{16}t^3 - \cdots Now, subtract this from 1: GX(t)=1(112t18t2116t3)G_X(t) = 1 - (1 - \frac{1}{2}t - \frac{1}{8}t^2 - \frac{1}{16}t^3 - \cdots) GX(t)=12t+18t2+116t3+G_X(t) = \frac{1}{2}t + \frac{1}{8}t^2 + \frac{1}{16}t^3 + \cdots

Step 3: Extract Probabilities P(X=k)P(X = k)

From the series expansion of GX(t)G_X(t), we can read off the probabilities: P(X=1)=12,P(X=2)=18,P(X=3)=116,P(X = 1) = \frac{1}{2}, \quad P(X = 2) = \frac{1}{8}, \quad P(X = 3) = \frac{1}{16}, \quad \dots In general, the probability P(X=k)P(X = k) is: P(X=k)=12kP(X = k) = \frac{1}{2^k} for k1k \geq 1.

Step 4: Why P(X=0)=0P(X = 0) = 0?

The probability P(X=0)P(X = 0) is the coefficient of t0t^0 in the series expansion of GX(t)G_X(t). From the expansion of GX(t)G_X(t), we see that the constant term (the coefficient of t0t^0) is 0. Thus: P(X=0)=0P(X = 0) = 0 This is because the generating function does not contain a constant term, which indicates that the probability of the random variable XX taking the value 0 is zero.

Conclusion

The probabilities P(X=k)P(X = k) for k=1,2,k = 1, 2, \dots are given by: P(X=k)=12kP(X = k) = \frac{1}{2^k} and P(X=0)=0P(X = 0) = 0 because there is no constant term in the generating function.

Would you like more details or have any questions?

Related Questions:

  1. What is the general form of a probability generating function?
  2. How can we derive the series expansion for other generating functions?
  3. What are the properties of random variables with generating functions of this form?
  4. How do we interpret the tail behavior of probabilities like P(X=k)P(X = k)?
  5. Can we find the expected value and variance of XX using its generating function?

Tip:

You can use the generating function to easily find moments like the expected value by differentiating it with respect to tt.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability Theory
Generating Functions
Series Expansion

Formulas

GX(t) = 1 - √(1 - t)
GX(t) = Σ P(X = k) t^k for k = 0 to ∞
√(1 - t) = Σ (-1)^n * (1/2 choose n) * t^n

Theorems

Binomial Series Expansion

Suitable Grade Level

Undergraduate Level (Probability/Statistics Course)