Let 𝑥1 and 𝑥2 be two independent R.V.S having a poisson distribution with parameters 𝜆1 and 𝜆2.

Find the probability distribution function of Y = 𝑥1 + 𝑥2.

Let \( X_1 \) and \( X_2 \) be two independent random variables (R.V.s), where \( X_1 \) follows a Poisson distribution with parameter \( \lambda_1 \) and \( X_2 \) follows a Poisson distribution with parameter \( \lambda_2 \). That is:

\[
X_1 \sim \text{Poisson}(\lambda_1)
\]
\[
X_2 \sim \text{Poisson}(\lambda_2)
\]

The probability mass function (PMF) of a Poisson-distributed random variable \( X \) with parameter \( \lambda \) is given by:

\[
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \quad k = 0, 1, 2, \dots
\]

Now, let \( Y = X_1 + X_2 \). We need to find the probability distribution function of \( Y \).

### Step 1: Determine the PMF of \( Y \).

Since \( X_1 \) and \( X_2 \) are independent, the probability generating function of \( Y = X_1 + X_2 \) can be obtained by the product of the generating functions of \( X_1 \) and \( X_2 \).

The probability generating function of a Poisson random variable \( X \) with parameter \( \lambda \) is given by:

\[
G_X(t) = \mathbb{E}(t^X) = e^{\lambda(t-1)}
\]

Thus, the generating function of \( Y \) is:

\[
G_Y(t) = G_{X_1}(t) \cdot G_{X_2}(t) = e^{\lambda_1(t-1)} \cdot e^{\lambda_2(t-1)} = e^{(\lambda_1 + \lambda_2)(t-1)}
\]

### Step 2: Identify the Distribution of \( Y \).

The generating function \( G_Y(t) = e^{(\lambda_1 + \lambda_2)(t-1)} \) corresponds to a Poisson distribution with parameter \( \lambda_1 + \lambda_2 \).

Therefore, \( Y \) follows a Poisson distribution with parameter \( \lambda_1 + \lambda_2 \). This means:

\[
Y \sim \text{Poisson}(\lambda_1 + \lambda_2)
\]

### Conclusion:

The probability distribution function of \( Y = X_1 + X_2 \) is:

\[
P(Y = k) = \frac{(\lambda_1 + \lambda_2)^k e^{-(\lambda_1 + \lambda_2)}}{k!}, \quad k = 0, 1, 2, \dots
\]

Would you like more details or have any questions?

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**Relative Questions:**
1. How does the independence of \( X_1 \) and \( X_2 \) affect the distribution of \( Y \)?
2. What would happen if \( X_1 \) and \( X_2 \) were not independent?
3. How can the result be generalized for the sum of more than two independent Poisson random variables?
4. Can the sum of two Poisson-distributed random variables ever be non-Poisson?
5. How does the convolution of probability distributions relate to this problem?

**Tip:** When dealing with sums of independent random variables, consider using generating functions or moment-generating functions to identify the resulting distribution.

Explore the probability distribution function of Y = X1 + X2, where X1 and X2 are independent Poisson-distributed random variables. Understand how the sum of independent Poisson random variables follows another Poisson distribution, derived using generating functions and probability mass functions (PMF).

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