Question

The number of withdrawals a bank processes in a day follows a random variable X. The number of deposits in a day is represented by Y. X and Y are independent and have the following moment generating functions \displaystyle {M}_{X}(t)=\frac{p}{1-q{e}^{t}} \displaystyle {M}_{Y}(t)={\left(\frac{p}{1-q{e}^{t}}\right)}^{r} q = 1 - p What is the expected number of transactions in a day?

Answer 1

Answer:

Check the explanation

Step-by-step explanation:

Number of transactions in a day is sum of number of withdrawals and number of deposits. So,

Number of transactions in a day, Z = X + Y

Moment Generating function of Z is,

T+1

Expected number of transactions in a day = E[Z]

$= \frac{\mathrm{d} }{\mathrm{d} t}_{t=0}M_Z(t) = (r+1)\left ( \frac{p}{1-qe^t} \right )^{r} * \frac{-p}{(1-qe^t)^2} * (-qe^t) for t = 0$

$= (r+1)\left ( \frac{p}{1-qe^0} \right )^{r} * \frac{-p}{(1-qe^0)^2} * (-qe^0)$

$= (r+1)\left ( \frac{p}{1-q} \right )^{r} * \frac{-p}{(1-q)^2} * (-q)$

$= (r+1)\left ( \frac{p}{p} \right )^{r} * \frac{-p}{(p)^2} * (-q)$

$= \frac{(r+1)q}{p}$