Expectation of Geometric random variable .

Question

For a discrete random variable a negative island exists- $$\Bbb E(X) = \sum_{x=0}^\infty \Bbb P(X\ |\!\!\!>x)$$

I need to use it, to calculate the Expectation of Geometric random variable.

Now, I don’t have any data about number of trials. So, how I do it? I know $x$ can be $1$, $2$, $3$, $\ldots$ and I know that the expectation of $x$ is $1/p$. So, how can I show it?

The question is bad written. "A negative island"? I understand that you should use the formula above, to calculate the expectation of a geometric random variable. So apply the formula: you need to calculate, for each $x$, which is the probability $\mathbb{P}(X | X>x)$ and then sum them up. — A. Pesare, Mar 14 '21 at 12:01

score 1 · Accepted Answer · answered Mar 14 '21 at 17:41

We know that $\mathbb P(X = n) = p(1-p)^{n-1}$ and from that we conclude that $$\mathbb P(X \leq n) = p \sum_{i=1}^n (1-p)^{i} = p \frac{1-(1-p)^n}{1-(1-p)} = 1-(1-p)^n$$ from the geometric series formula, namely $$\sum_{i=0}^{n-1} q^i = \frac{1-q^n}{1-q}.$$

Now we can conclude that $\mathbb P(X >n) = 1- \mathbb P(X \leq n) = (1-p)^n$. Now we use the given formula to calculate the expected value, using $(1-p) < 1$ (because if $p=0$ we don't have an interesting distribution anyway, therefore we have $p>0$) and the formula for the geometric series (where the $q^n$-part goes to zero because $|q|<1$ in our case): $$\mathbb E[X] = \sum_{x=0}^{\infty} \mathbb P(X >x) = \sum_{x=0}^{\infty} (1-p)^x = \frac{1}{1-(1-p)} = \frac{1}{p}$$

Expectation of Geometric random variable .

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