Question

Suppose that $X$takes on one of the values$0,1$and$2$. If for some constant$c,P\{X=i\}=cP\{X=i-1\},i=1,2$, find$E\left[X\right].$

Short answer

The value of$E\left(X\right)=\frac{c}{1+c+c^2}[1+2c]$.

Step-by-step solution

Given Information

Given that a random variable $X$takes values $0,1,$and 2. For some constant $c,$

$P[X=i]=P[X=i-1],i=1,2$

Substitute the Value

Substitute ${}^{i=1}$in equation (1),

$P[X=1]=cP[X=0]\dots ..\left(2\right)$

Now, substitute $i=2$in equation (1),

$P[X=2]=cP[X=1]$

Substitute the value of $P[X=1]$using equation

(2) in $P[X=2]$.

Thus,

$P[X=2]=c[cP[X=0\left]\right]$

$P[X=2]=c^{2}P[X=0]$

Calculation of the Value

Suppose that$P[X=0]=p.$

Substitute $P[X=0]$in equations $\left(2\right)$and $\left(3\right).$

$P[X=1]=cp$.........$\left(1\right)$

$P[X=2]=c^{2}p$.........$\left(2\right)$

Since $X$is a random variable and takes a value $0$, Since $X$is a random variable and takes value $0,1,2,$than by the property of probability mass function, $P[X=0]+P[X=1]+P[X=2]=1.$

Thus,

$p+cp+c{p}^2=1$

$\Rightarrow p\left[1+c+c^2\right]=1$

We get,

$\Rightarrow p=\frac{1}{1+c+c^2}$

Computation of Expectation of X

Therefore, on substituting the value of $p$in equation $\left(4\right)$and$\left(5\right)$we have,

$P[X=1]=\frac{c}{1+c+c^2}$

$P[X=2]=\frac{c^2}{1+c+c^2}$

$P[X=0]=\frac{1}{1+c+c^2}$

Compute expectation of $X$as follows:

$E\left(X\right)=\sum xP[X=x]$

$=0\left[\frac{1}{1+c+c^2}\right]+1\left[\frac{c}{1+c+c^2}\right]+2\left[\frac{c^2}{1+c+c^2}\right]$

$=\frac{c}{1+c+c^2}+\frac{2c^2}{1+c+c^2}$

We get$=\frac{c}{1+c+c^2}[1+2c].$

Final Answer

The value of $E\left(X\right)$$=\frac{c}{1+c+c^2}[1+2c]$