91Ó°ÊÓ

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$P\{X+Y=2\}=?$

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

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

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

$=P[X=0]â‹\dots P[Y=2]+P[X=1]â‹\dots P[Y=1]+P[X=2]â‹\dots P[Y=0]$

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

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

localid="1647493526857" $=e^{−2}\left[4.451074×{10}^{−4}\right]$

$=6.024×{10}^{−5}$

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

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

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

Then role="math" localid="1647493957328" $P[XY=0]=?$?

b) $P[XY=0]=P[X=0$orrole="math" localid="1647494461034" $Y=0]$

role="math" localid="1647494559363" $P[X=0]+P[Y=0]−P[X=0,Y=0]$

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

role="math" localid="1647494541691" $=0.1353+0.000000954−0.000000129$

$=0.135300825$

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

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$E[XY]=?$

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

$=\left(\mathrm{λ}\right)·\left(np\right)$

$=\left(2\right)â‹\dots (10×0.75)$

$=15$

Hence, the value is$E[X,Y]=15$