Question

If a discrete random variable X is defined as follows P [ X = x ] = { k ( x + 1 ) 5 x , if x = 0 , 1 , 2 … … 0 , otherwise

Answer 1

k = 16/25

Answer 2

We are given the probability mass function (pmf) of a discrete random variable $X$ as

$$P[X=x] = \begin{cases} k\,(x+1) \, 5^{-x} & \text{for } x = 0,1,2,\ldots, \\ 0 & \text{otherwise.} \end{cases}$$

For $P[X=x]$ to be a valid pmf, we need

$$\sum_{x=0}^{\infty} P[X=x] = \sum_{x=0}^{\infty} k\,(x+1) \, 5^{-x} = 1.$$

Step 1. Factor out $k$:

$$k \sum_{x=0}^{\infty} (x+1) \, 5^{-x} = 1.$$

Step 2. Evaluate the sum

$$S = \sum_{x=0}^{\infty} (x+1) \, r^x,$$

with $r = \frac{1}{5}$. It is known that

$$\sum_{x=0}^{\infty} (x+1)\,r^x = \frac{1}{(1-r)^2} \quad \text{for } |r|<1.$$

Substitute $r=\frac{1}{5}$:

$$S = \frac{1}{\left(1-\frac{1}{5}\right)^2} = \frac{1}{\left(\frac{4}{5}\right)^2} = \frac{1}{\frac{16}{25}} = \frac{25}{16}.$$

Step 3. Now, putting it back into the normalization condition:

$$k \cdot \frac{25}{16} = 1 \quad \Longrightarrow \quad k = \frac{16}{25}.$$