Finding Var(X) from a distribution

Question

I saw this question from an AQA exam:-

The discrete random variable $X$ has probability distribution given by

$P(X = x) = \dfrac {e^{-\lambda}\lambda ^x}{x!}$ for x = 0, 1, 2, … or 0 otherwise.

Find Var(X).

My questions are:-

(1) When I see the probability distribution, I should immediately recognize (assume) that it is a POISSON distribution even the question did not mention that explicitly? (Y/N)

(2) To find Var (X) in general, I should use the standard formula $Var (X) = E(X^2) – (E(X))^2$. But if it is a POISSON distribution, I should use the formula $Var(X) = E(X(X-1))$ instead (meaning the mentioned standard formula should NOT be used at all)? (Y/N)

(3) Can $Var(X) = E(X(X-1))$ be derived from $Var (X) = E(X^2) – (E(X))^2$ for all distributions; or the second is true only when the distribution is POISSON? And if it is derivable please give me a link.

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Sorry for reading the marking scheme wrongly. Instead of re-writing my original post, I include the correction in here:-

The question has the first part "Show that $E(X) = \lambda$".

To find Var(x), we need to find $E(X(X - 1)) = ... = \lambda ^2$ first.

Next, from the standard formula, $Var (X) = E(X(X - 1)) + E(X) - (E(X))^2 = E(X(X - 1)) + \lambda - \lambda ^2 = \lambda$.

My further question is then "in finding Var(X), is introducing the x(x-1) factor to the summand in the summation a standard trick?"

Answer 1

No need to assume anything, you can use the basic expression $Var(X) = E[X^2] - E[X]^2$ and solve the question. It happens to be Poisson distribution so you can directly write the variance as $\lambda$ as well.

I have no idea where you got the expression $Var(X) = E[X(X-1)]$. It is simply wrong. $$E[X(X-1)] = E[X^2]-E[X]$$ $$Var(X) = E[X^2]-E[X]^2 $$ So obviously for a general case, $$ Var(X) \neq E[X(X-1)]$$

Answer 2

I think you are confusing a method to easily compute the variance with the actual value of the variance. What I mean is that it is easier to directly compute $$\operatorname{E}[X(X-1)] = \sum_{x=0}^\infty x(x-1) e^{-\lambda} \frac{\lambda^x}{x!} = \lambda^2 \sum_{x=2}^\infty e^{-\lambda} \frac{\lambda^{x-2}}{(x-2)!} = \lambda^2,$$ rather than $$\operatorname{E}[X^2] = \sum_{x=0}^\infty x^2 e^{-\lambda} \frac{\lambda^x}{x!}.$$ In the first calculation, we observe that the summand is zero for $x \in \{0,1\}$, which is why we can ignore the first two terms. We also note that for $x \ge 2$, $x! = x(x-1)(x-2)!$. Then pulling out a factor of $\lambda^2$, we obtain the sum of the same Poisson distribution over its support; thus it equals $1$, and the expectation $\operatorname{E}[X(X-1)]$ is $\lambda^2$. Meanwhile, the second computation is not so simple because we do not have a convenient cancellation of the second factor of $x$, and it is not obvious how to proceed.

The first calculation is in fact a specific case of the more general identity $$\operatorname{E}[X(X-1)\cdots (X-k+1)] = \lambda^k, \quad k \in \{1, 2, \ldots\}.$$

We may then use this calculation to obtain the desired variance; i.e., $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \operatorname{E}[X(X-1) + X] - \operatorname{E}[X]^2 + \operatorname{E}[X(X-1)] + \operatorname{E}[X] - \operatorname{E}[X]^2,$$ hence $$\operatorname{Var}[X] = \lambda^2 + \lambda - \lambda^2 = \lambda.$$