question_answer

The random variable X can take only the values 0, 1, 2. Given that \[P(X=0)=P(X=1)=p\] and that \[E({{X}^{2}})=E(X),\] find the value of p.

Answer:

Clearly, \[P(X=0)+P(X=1)+P(X=2)=1\]             \[\Rightarrow \]   \[p+p+P(X=2)=1\]             \[\Rightarrow \]   \[P(X=2)=1-2p\] So, the probability distribution of X is as given below

\[{{\mathbf{x}}_{\mathbf{i}}}\]	0	1	2
\[{{\mathbf{p}}_{\mathbf{i}}}\]	p	p	\[(1-2p)\]

            \[\therefore \] \[E(X)=0\times p\,+1\times p+2(1-2p)=2-3p\]             and \[E({{X}^{2}})={{0}^{2}}\times p\,+{{1}^{2}}\times p+{{2}^{2}}(1-2p)=4-7p\]             It is given that,                         \[E({{X}^{2}})=E(X)\] \[\Rightarrow \] \[4-7p=2-3p\]       \[\Rightarrow \]  \[4p=2\]            \[\Rightarrow \] \[p=\frac{1}{2}\]