The random variable X can take only the values 0, 1, 2. Given tha… - Vidyadip

Question

The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P (X = 1) = p and that E(X²) = E[X], find the value of p.

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Answer

Since, X = 0, 1, 2 and P (X) at X = 0 and 1 is p, let at X = 2, P (X) is x.
⇒ p + p + x = 1
⇒ x = 1 – 2p
We get, the following distribution

X	0	1	2
P(X)	p	q	1 - 2p

$\therefore\text{E}[\text{X}]=\sum\text{XP}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+2(1-2\text{p})$
$=\text{p}+2-4\text{p}$
$=2-3\text{p}$
And $\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+4\cdot(1-2\text{p})$
$=\text{p}+4-8\text{p}$
$=2-7\text{p}$
Also, given that $\text{E}(\text{X}^2)=\text{E}[\text{X}]$
$\Rightarrow4-7\text{p}=2-3\text{p}$
$\Rightarrow4\text{p}=2$
$\Rightarrow\text{p}=\frac{1}{2}$

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