hroncits8y

2021-11-29

Suppose that the random variable X is defined as $X=0.6{X}_{1}+0.3{X}_{2}+0.1{X}_{3}$, where the probability distributions of the independent random variables ${X}_{1},{X}_{2},\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{X}_{3}$ are
${f}_{X1}\left(x\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.1}^{x}×{0.9}^{5-x}$
${f}_{X2}\left(x\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.4}^{x}×{0.6}^{5-x}$
${f}_{X3}\left(3\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.6}^{x}×{0.4}^{5-x}$
$x=0,1,2,3,4,5$
a) Evaluate the mean of X.
b) Evaluate the variance of X.

Archie Griffin

Step 1
Given:
$X=0.6{X}_{1}+0.3{X}_{2}+0.1{X}_{3}$
Where all three are independent
${f}_{X1}\left(x\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.1}^{x}×{0.9}^{5-x}$
${f}_{X2}\left(x\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.4}^{x}×{0.6}^{5-x}$
${f}_{X3}\left(3\right)=\left(\begin{array}{c}5\\ x\end{array}\right){0.6}^{x}×{0.4}^{5-x}$
$x=0,1,2,3,4,5$
Step 2
It can be seen that all the three distributions follows binomial distribution
So, Mean of binomial is
$n×p$
Variance of binomial is
$n×p×q$
So, ${X}_{1}\sim B\in \left(5,0.1\right)$
$E\left[{X}_{1}\right]=5×0.1=0.5$
$Var\left[{X}_{1}\right]=5×0.1×0.9=0.45$
${X}_{2}\sim B\in \left(5,0.4\right)$
$E\left[{X}_{2}\right]=5×0.4=2$
$Var\left[{X}_{2}\right]=5×0.4×0.6=1.2$
${X}_{3}\sim B\in \left(5,0.6\right)$
$E\left[{X}_{3}\right]=5×0.6=3$
$Var\left[{X}_{3}\right]=5×0.6×0.4=1.2$
Step 3
a) Mean of X
$E\left[X\right]=0.6×E\left[{X}_{1}\right]+0.3×E\left[{X}_{2}\right]+0.1×E\left[{X}_{3}\right]$
$E\left[X\right]=0.6×0.5+0.3×2+0.1×3$
$E\left[X\right]=1.2$
Hence, the mean is 1.2
Step 4
b) If the events are independent, then the variance is
$Var\left[A+B\right]=Var\left[A\right]+Var\left[B\right]$
Here, all are independent, so the variance of X can be calculated as follows
$Var\left[X\right]={0.6}^{2}×Var\left[{X}_{1}\right]+{0.3}^{2}×Var\left[{X}_{2}\right]+{0.1}^{2}×Var\left[{X}_{3}\right]$
$Var\left[X\right]={0.6}^{2}×0.45+{0.3}^{2}×1.2+{0.1}^{2}×1.2$

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