Q.7.75 - A First Course in Probability

Question

The moment generating function of $X$ is given by $M_X\left(t\right)=\exp \{2e^t-2\}$nd that of $Y$ by $M_Y\left(T\right)={(\frac{3}{4}{e}^t+\frac{1}{4})}^{10}$. If $X$ and $Y$ are independent, what are

(a) $P\{X+Y=2\}$?

(b) $P\{XY=0\}$?

(c) $E[X Y]$?

Step-by-Step Solution

Answer

a) The value of $P\{X+Y=2\}$ is $6.024\times {10}^{-5}$

b) The value of $\mathrm{P}[\mathrm{XY}=0]$ is $0.135300825$

c) The value of $E\left[XY\right]$ is $15$

Step 1: Given Information (Part a)

Function of $X=M_X\left(t\right)=\exp \left\{2e^t-2\right\}$

Function of $Y=M_Y\left(T\right)={\left(\frac{3}{4}{e}^t+\frac{1}{4}\right)}^{10}$

$(X,Y)=$Independent,

Then $P\{X+Y=2\}=?$

Step 2: Explanation (Part a)

${\mathrm{M}}_{\mathrm{X}}\left(\mathrm{t}\right)=\exp \left\{2\left({\mathrm{e}}^{\mathrm{t}}-1\right)\right\}\Rightarrow \mathrm{X}~$Poisson

${\mathrm{M}}_{\mathrm{Y}}\left(\mathrm{t}\right)={\left(\frac{3}{4}{e}^t+\frac{1}{4}\right)}^{10}\Rightarrow Y~\text{ Binomial }\left(10,\frac{3}{4}\right)$

Calculate $P\{X+Y=2\}$

$\mathrm{P}[\mathrm{X}+\mathrm{Y}=2]=\mathrm{P}[\mathrm{X}=0,\mathrm{Y}=2]+\mathrm{P}[\mathrm{X}=1,\mathrm{Y}=1]+\mathrm{P}[\mathrm{X}=2,\mathrm{Y}=0]$

Here $(X,Y)$ are independent, so

$=P[X=0]\cdot P[Y=2]+P[X=1]\cdot P[Y=1]+P[X=2]\cdot P[Y=0]$

$\begin{array}{r}=\left(e^{-2}\right)\left(0.000386\right)+\left(2e^{-2}\right)\left(0.0000286\right)+\left(\frac{2^2{e}^{-2}}{2!}\right)\left(9.537\times {10}^{-7}\right)\end{array}$

$=e^{-2}\left[0.000386+\left(5.72\times {10}^{-5}\right)+\left(1.9074\times {10}^{-6}\right)\right]$

$=e^{-2}\left[4.451074\times {10}^{-4}\right]$

$=6.024\times {10}^{-5}$

Step 3: Final Answer (Part a)

Hence, the value of$P\{X+Y=2\}$ is $6.024\times {10}^{-5}.$

Step 1: Given Information (Part b)

Function of $X=M_X\left(t\right)=\exp \left\{2e^t-2\right\}$

Function of $Y=M_Y\left(T\right)={\left(\frac{3}{4}{e}^t+\frac{1}{4}\right)}^{10}$

$(X, Y)=$Independent,

Then $P[X Y=0]=?$?

Step 2: Explanation (Part b)

b) $P[XY=0]=P[X=0$ or $Y=0]$

$P[X=0]+P[Y=0]-P[X=0,Y=0]$

$\begin{array}{l}=e^{-2}+(9.5367\times {10}^{-7})-\left[e^{-2}(9.5367\times {10}^{-7})\right]\end{array}$

$=0.1353+0.000000954-0.000000129$

$=0.135300825$

Step 3: Final Answer (Part b)

Hence, the value is $P[XY=0]=0.135300825$

Step 1:

Function of $X=M_X\left(t\right)=\exp \left\{2e^t-2\right\}$

Function of $Y=M_Y\left(T\right)={\left(\frac{3}{4}{e}^t+\frac{1}{4}\right)}^{10}$

$(X, Y)=$Independent,

Then $E[X Y]=?$

Step 2: Explanation (Part c)

c) $\mathrm{E}\left[\mathrm{XY}\right]=\mathrm{E}\left[\mathrm{X}\right]\cdot \mathrm{E}\left[\mathrm{Y}\right]$$(\mathrm{X}$ and $Y)$ are independent

$=\left(\lambda \right)·\left(np\right)$

$=\left(2\right)\cdot (10\times 0.75)$

$=15$

Step 9: Final Answer (c)

Hence, the value is $E[X,Y]=15$