 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.

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

### Step 1: Generating Function and Definition

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

### Step 2: Expand the Generating Function

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

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

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

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

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

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

### Conclusion

The probabilities \( P(X = k) \) for \( k = 1, 2, \dots \) are given by:
\[
P(X = k) = \frac{1}{2^k}
\]
and \( P(X = 0) = 0 \) 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 \( P(X = k) \)?
5. Can we find the expected value and variance of \( X \) 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 \( t \).

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.

Learn how to find probabilities P(X = k) for k = 1, 2, ... using the generating function GX(t) = 1 − √(1 − t). This solution involves expanding the generating function as a power series and extracting coefficients for each probability. Suitable for undergraduate students studying probability theory.

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