Question

Let \( X_1, X_2, X_3 \) and \( X_4 \) be i.i.d. discrete random variables with the probability mass function 

\[ P(X_1 = n) = \begin{cases} \dfrac{3^{n-1}}{4^n}, & n = 1, 2, \dots, \\ 0, & \text{otherwise}. \end{cases} \] 

Then \( P(X_1 + X_2 + X_3 + X_4 = 6) \) equals ............... 
 

Hint:

For sums of i.i.d. random variables, use convolution to find the probability of their sum.

Answer 1

Correct Answer: 0.01 - 0.03

Step 1: Verify this is a valid PMF

$$\sum_{n=1}^{\infty} P(X_1 = n) = \sum_{n=1}^{\infty} \frac{3^{n-1}}{4^n} = \sum_{n=1}^{\infty} \frac{3^{n-1}}{4^n} = \frac{1}{4} \sum_{n=1}^{\infty} \left(\frac{3}{4}\right)^{n-1} = \frac{1}{4} \cdot \frac{1}{1-3/4} = \frac{1}{4} \cdot 4 = 1$$ 

Step 2: Find $P(X_1 + X_2 + X_3 + X_4 = 6)$

We need to find all ways to write $6 = n_1 + n_2 + n_3 + n_4$ where each $n_i \geq 1$.

The possible partitions of 6 into 4 positive integers are:

$(1,1,1,3)$ and permutations

$(1,1,2,2)$ and permutations

$(1,2,2,1)$ is already counted above

$(2,2,1,1)$ is already counted above

$(1,1,4,0)$ - not valid (need positive integers)

Wait, let me be systematic. We need ordered tuples $(n_1, n_2, n_3, n_4)$ where each $n_i \geq 1$ and $n_1 + n_2 + n_3 + n_4 = 6$.

Using stars and bars, this is equivalent to distributing $6-4=2$ extra items among 4 variables.

The compositions are:

(1,1,1,3): permutations = $\frac{4!}{3!1!} = 4$ ways

(1,1,2,2): permutations = $\frac{4!}{2!2!} = 6$ ways

(1,2,1,2): already counted in (1,1,2,2)

Let me list them systematically:

Type 1: Three 1's and one 3: $(1,1,1,3), (1,1,3,1), (1,3,1,1), (3,1,1,1)$ → 4 ways

Type 2: Two 1's and two 2's: $(1,1,2,2), (1,2,1,2), (1,2,2,1), (2,1,1,2), (2,1,2,1), (2,2,1,1)$ → 6 ways

Total: 10 ways

Step 3: Calculate probabilities

For Type 1: $(1,1,1,3)$ $$P(X_1=1, X_2=1, X_3=1, X_4=3) = P(X_1=1)^3 \cdot P(X_1=3)$$ $$= \left(\frac{3^0}{4^1}\right)^3 \cdot \frac{3^2}{4^3} = \left(\frac{1}{4}\right)^3 \cdot \frac{9}{64} = \frac{1}{64} \cdot \frac{9}{64} = \frac{9}{4096}$$

For Type 2: $(1,1,2,2)$ $$P(X_1=1, X_2=1, X_3=2, X_4=2) = P(X_1=1)^2 \cdot P(X_1=2)^2$$ $$= \left(\frac{1}{4}\right)^2 \cdot \left(\frac{3}{16}\right)^2 = \frac{1}{16} \cdot \frac{9}{256} = \frac{9}{4096}$$

Step 4: Total probability

$$P(X_1 + X_2 + X_3 + X_4 = 6) = 4 \cdot \frac{9}{4096} + 6 \cdot \frac{9}{4096}$$ $$= 10 \cdot \frac{9}{4096} = \frac{90}{4096} = \frac{45}{2048}$$

Step 5: Convert to decimal

$$\frac{45}{2048} \approx 0.021973$$

Answer

$$P(X_1 + X_2 + X_3 + X_4 = 6) = \frac{45}{2048} \approx 0.022$$