Question 1160924: Let X be a discrete random variable with PMF
Fx(k)={c/3^k For k=3,4,5...
{0 Otherwise
Find the value of Constant C.
Also, find E(X),
E(X^2) and
Var(X).
Answer by CPhill(2285)   (Show Source):
You can put this solution on YOUR website! ```python
import sympy as sp
k, c = sp.symbols('k c')
# Sum of c / 3^k for k=3 to infinity
sum_pmf = sp.summation(c / (3**k), (k, 3, sp.oo))
c_val = sp.solve(sum_pmf - 1, c)[0]
print(f"c_val = {c_val}")
# Expected value E(X) = sum_k=3^\infty k * (c / 3^k)
EX = sp.summation(k * (c_val / (3**k)), (k, 3, sp.oo))
EX2 = sp.summation((k**2) * (c_val / (3**k)), (k, 3, sp.oo))
VarX = EX2 - EX**2
print(f"E(X) = {EX}")
print(f"E(X^2) = {EX2}")
print(f"Var(X) = {VarX}")
```
```text
c_val = 18
E(X) = 7/2
E(X^2) = 13
Var(X) = 3/4
```
Here is the step-by-step solution to find the constant $c$, the expected values $E(X)$ and $E(X^2)$, and the variance $\text{Var}(X)$.
---
### Step 1: Find the Value of Constant $c$
The sum of all probabilities over the support of $X$ ($k = 3, 4, 5, \dots$) must equal $1$:
$$\sum_{k=3}^{\infty} P(X = k) = 1 \implies \sum_{k=3}^{\infty} \frac{c}{3^k} = 1$$
Factor out $c$ and use the geometric series formula $\sum_{k=n}^{\infty} r^k = \frac{r^n}{1 - r}$:
$$c \sum_{k=3}^{\infty} \left(\frac{1}{3}\right)^k = c \cdot \left( \frac{(1/3)^3}{1 - 1/3} \right) = 1$$
$$c \cdot \left( \frac{1/27}{2/3} \right) = 1$$
$$c \cdot \left( \frac{1}{18} \right) = 1 \implies \mathbf{c = 18}$$
Thus, the probability mass function is:
$$P(X = k) = \frac{18}{3^k} = 2 \cdot 3^{2-k} \quad \text{for } k = 3, 4, 5, \dots$$
---
### Step 2: Find the Expected Value $E(X)$
The expected value is defined as:
$$E(X) = \sum_{k=3}^{\infty} k \cdot P(X = k) = \sum_{k=3}^{\infty} k \cdot \frac{18}{3^k}$$
Let $j = k - 2$, so when $k = 3$, $j = 1$:
$$E(X) = \sum_{j=1}^{\infty} (j + 2) \cdot \frac{18}{3^{j+2}} = \sum_{j=1}^{\infty} (j + 2) \cdot 2 \cdot \left(\frac{1}{3}\right)^j$$
Using the formula for the mean of a Geometric distribution with success probability $p = 1 - 1/3 = 2/3$ shifted by $2$:
Alternatively, computing the infinite series directly yields:
$$E(X) = 2 \sum_{j=1}^{\infty} j \left(\frac{1}{3}\right)^j + 4 \sum_{j=1}^{\infty} \left(\frac{1}{3}\right)^j$$
* First series: $\sum_{j=1}^{\infty} j \left(\frac{1}{3}\right)^j = \frac{1/3}{(1 - 1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$
* Second series: $\sum_{j=1}^{\infty} \left(\frac{1}{3}\right)^j = \frac{1/3}{1 - 1/3} = \frac{1}{2}$
$$E(X) = 2\left(\frac{3}{4}\right) + 4\left(\frac{1}{2}\right) = \frac{3}{2} + 2 = \mathbf{\frac{7}{2}} = 3.5$$
---
### Step 3: Find $E(X^2)$
$$E(X^2) = \sum_{k=3}^{\infty} k^2 \cdot \frac{18}{3^k}$$
Using $k = j + 2$ for $j = 1, 2, 3, \dots$:
$$E(X^2) = \sum_{j=1}^{\infty} (j+2)^2 \cdot \frac{18}{3^{j+2}} = 2 \sum_{j=1}^{\infty} (j^2 + 4j + 4) \left(\frac{1}{3}\right)^j$$
Evaluating the series sums:
* $\sum_{j=1}^{\infty} j^2 \left(\frac{1}{3}\right)^j = \frac{(1/3)(1 + 1/3)}{(1 - 1/3)^3} = \frac{4/9}{8/27} = \frac{3}{2}$
* $\sum_{j=1}^{\infty} j \left(\frac{1}{3}\right)^j = \frac{3}{4}$
* $\sum_{j=1}^{\infty} \left(\frac{1}{3}\right)^j = \frac{1}{2}$
Combine the results:
$$E(X^2) = 2 \left( \frac{3}{2} + 4\left(\frac{3}{4}\right) + 4\left(\frac{1}{2}\right) \right) = 2 \left( \frac{3}{2} + 3 + 2 \right) = 2 \left( \frac{13}{2} \right) = \mathbf{13}$$
---
### Step 4: Find the Variance $\text{Var}(X)$
$$\text{Var}(X) = E(X^2) - [E(X)]^2$$
$$\text{Var}(X) = 13 - \left(\frac{7}{2}\right)^2 = 13 - \frac{49}{4} = \frac{52 - 49}{4} = \mathbf{\frac{3}{4}} = 0.75$$
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