Expectation of a non negative integer value random variable

Question

the expectation for a discrete type random variable that takes non negative integer value is given by:

$$ E[X] = \sum_{k=0}^\infty kP(X=k) = \sum_{k=0}^\infty P(X>k) $$

The derivation involves the following property: $$ P(X>n) = \sum_{k=n}^\infty P(X=k) $$

in a post on this website (https://stats.stackexchange.com/questions/476887/intuition-for-expectation-of-discrete-random-variable-that-takes-positive-intege) ,enter image description herethe derivation of the upper mentioned property is done in the following way: $$ P(X>n) = 1 - P(X\leq n) = \sum_{k=0}^\infty P(X=k)\ - \sum_{k=n}^\infty P(X=k) = \sum_{k=n}^\infty P(X=k) $$

but I think this is incorrect because there should be equality sign in $ P(X>n)$ because we are also considering the case of $P(X=n)$ in the summation:

as a consequence of this the expectation formulae would also involve the equality i.e. $$ E[X] = \sum_{k=0}^\infty P(X\geq k) $$

Is it correct or does the equality doesn't matter and if it does what consequence does this have ? I have also attached a image of my version of derivation please tell me if there's any mistake.

Answer 1

There seems to be some typo in the indexes, so if you are good at double sum, just go ahead like this:

$$ \begin{align} E[X] &= \sum_{k=0}^{\infty}k\Pr\{X = k\} \\ &= \sum_{k=1}^{\infty}k\Pr\{X = k\} \\ &= \sum_{k=1}^{\infty}\left(\sum_{i=1}^k 1\right)\Pr\{X = k\} \\ &= \sum_{k=1}^{\infty}\sum_{i=1}^k \Pr\{X = k\} \\ &= \sum_{i=1}^{\infty}\sum_{k=i}^{\infty}\Pr\{X = k\} \\ &= \sum_{i=1}^{\infty}\Pr\{X \geq i\} \end{align}$$

The key at interchanging the double sum is considering that as the following inequality for the indices: $$ 1 \leq i \leq k < \infty $$

So depends on which index is being used first in the inner sum, we will use the corresponding bound for that index.

Answer 2

Consider the term $0P(X=0)$ in the standard formula for the expectation. Compare that to the terms $P(X>0)$ and $P(X\geq 0)$ in the alternative formulas. See that it must be $X>0$.

More concretely, apply the given theorem and your alternative theorem on some really simple distribution, like the constant random variable with $P(X=3)=1$, or a coin toss $P(X=0)=P(X=1)=\frac12$. See which one gives the right result and why.