Question

If X is a random passion variate such that $E({X^2}) = 6$, then $E(X) =$
A.-3
B.2
C.-3 & 2
D.-2

Answer 1

Let $x$ will be the average number of the occurrence of success events in a particular time in the poisson distribution. Then the mean and variance of the variable are equal i.e.
$\Rightarrow Var(X) = E(X) = x$
And put this value in the formula of variance which is given by
$\Rightarrow Var(X) = E({X^2}) - {\left[ {E(X)} \right]^2}$

Complete step-by-step answer:
The probability distribution in which probability of a variable expressed the events occurring in a fixed interval of time or space and these events occur with a constant mean rate is known as poison distribution and it is a discreate probability distribution.
In this question let us see what is given? We are given with the value of $E({X^2})$i.e.
$\Rightarrow E({X^2}) = 6$ ………..(1)
And we have to find the value of $E(X)$i.e. mean of the variable.
First of all, let us assume the value of $E(X)$be $x$i.e.$x$
$\Rightarrow E(X) = x$ …………(2)
In the poison distribution the mean and variance of variable X both are equal i.e.
$\Rightarrow Var(X) = E(X)$ ………….(3)
From (2) and (3) we get,
$\Rightarrow Var(X) = x$ ………….(4)
variance is given by expected value of the squared difference between the variable and its expected value i.e.
$\Rightarrow Var\left( X \right) = E\left( {{{\left[ {X - E\left( X \right)} \right]}^2}} \right)$
By simplifying it we get,
$\Rightarrow Var\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$ …………(5)
Put the value of (1), (2) and (3) in equation (5), we get,
$\Rightarrow x = 6 - {\left[ x \right]^2}$
By opening the bracket, we get,
$\Rightarrow x = 6 - {x^2}$
Taking all the terms on L.H.S, we get,
$\Rightarrow {x^2} + x - 6 = 0$
Solve the above quadratic equation by finding factors of 6 i.e. 2 and 3 and when we take this combination and by subtracting them, we get 1. Hence, we can write the term of $x$as $x = 3x - 2x$and putting the value in the above equation, we will get,
$\Rightarrow {x^2} + 3x - 2x - 6 = 0$
Taking common 3 from first two terms and -2 from last two terms we get,
$\Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right) = 0$
Taking $x + 3$ common from both terms, we get,
$\Rightarrow \left( {x - 2} \right)\left( {x + 3} \right) = 0$
Therefore, $x - 2 = 0$ and $x + 3 = 0$
Hence, $x = 2$ and $x = - 3$
Therefore, the correct answer is option C.

Note: Mean and variance of poisson distribution:
We can prove that in poisson distribution mean and variance are equal.
The probability mass function i.e. pmf of the poisson distribution is given by
$\Rightarrow P\left( x \right) = \dfrac{{{\lambda ^x}.{e^{ - \lambda }}}}{{x!}}$, where $x$is 0,1,2,3…….$\infty$
Mean i.e. $E(x)$ can be calculated as
$\Rightarrow E\left( x \right) = \sum\limits_{x = 0}^\infty {xP\left( x \right)}$
Put the value of $P\left( x \right)$ in this equation, we get,
$\Rightarrow E\left( x \right) = \sum\limits_{x = 0}^\infty {x\dfrac{{{\lambda ^x}.{e^{ - \lambda }}}}{{x!}}}$
By solving this equation, we will get,
$\Rightarrow E\left( x \right) = \lambda$
Now find $E\left( {x\left( {x - 1} \right)} \right)$ and we will get this value of this i.e.
$\Rightarrow E\left( {x\left( {x - 1} \right)} \right) = {\lambda ^2}$
Now, $E\left( {{x^2}} \right) = E\left( {x\left( {x - 1} \right)} \right) + E\left( x \right)$ and putting the value of $E\left( {x\left( {x - 1} \right)} \right)$ and $E(x)$ in this equation we get,
$\Rightarrow E\left( {{x^2}} \right) = {\lambda ^2} + \lambda$
As we know,
$\Rightarrow Var\left( X \right) = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}$
And putting the value of $E(x)$ and $E\left( {{x^2}} \right)$, we get,
$\Rightarrow Var\left( X \right) = {\lambda ^2} + \lambda - {\left( \lambda \right)^2}$
Hence, $Var\left( X \right) = \lambda$
Therefore, in poisson distribution mean and variance are equal. Hence, proved.