Question

Suppose that $P\{X=0\}=1-P\{X=1\}$. If $E\left[X\right]=3Var\left(X\right)$, find $P\{X=0\}$.

Short answer

Observe two cases. If $EX=0$, the answer is $p=1$. Otherwise, answer$\frac{1}{3}$.

Step-by-step solution

Given information

Given in the question is a function,

$P\{X=0\}=1-P\{X=1\}$. If $E\left[X\right]=3Var\left(X\right)$.

Computation

Define $P(X=0)=p$.

Then we have that $P(X=1)=1-p$.

From the given relation between the mean and variance, we have that

localid="1646920607051" $E\left(X\right)=3Var\left(X\right)=3E\left(X^2\right)-3{\left(EX\right)}^2=3EX-3(EX{)}^2$

$\Rightarrow 3(EX{)}^2=2EX$.

Calculation

The last inequality in the first row holds, since $X$only assumes values $0$and $l$, so $X~X^2$.

If localid="1646920627387" $E\left(X\right)=0$, we have that $p=1$. So, suppose that localid="1646920639456" $E\left(X\right)>0$. Then, we have that

localid="1646920651370" $E\left(X\right)=\frac{2}{3}$

Hence,

localid="1646920661162" $\frac{2}{3}=E\left(X\right)=1-p\Rightarrow p=\frac{1}{3}$.

Final answer

We have observe two cases.

If $E\left(X\right)=0$, the answer is $p=1$. Otherwise, answer is$1/3$.