Deriving First Success Expectation with linearity property of expectation?

Question

I'm a newbie studying probability.

Let $X$ be random variables distributed as Geometric Distribution: $X \sim \text{Geom}(p)$.

Let $Y$ be random variables distributed as First Success Distribution: $Y \sim \text{FS}(p)$.

I know how to calculate the Geometric expectation:

$$E(X) = \sum_{k=0}^{\infty}kq^kp = pq\sum_{k=0}^{\infty}kq^{k-1} = pq\frac{1}{(1-q)^2} = \frac{p}{q}$$

and the First Success expectation: $$E(Y) = \sum_{k=1}^{\infty}kq^{k-1}p = p\sum_{k=1}^{\infty}kq^{k-1} = p\frac{1}{(1-q)^2} = \frac{1}{p}$$

But in the book there is another way to derive the formula for $E(Y)$.

it says: "Since we can write $Y \sim \text{FS}(p)$ as $Y = X + 1$ where $X \sim \text{Geom}(p)$, we have $E(Y) = E(X+1)=\frac{q}{p}+1=\frac{1}{p}$."

I know that both method arrive at the same formula but why the second derivation is valid?

is it because of the linearity property of the expectation? such as $E(X+Y) = E(X) + E(Y)$ and in this case it is $E(Y) = E(X+1) = E(X) + E(1) = \frac{q}{p}+1.$ I didn't know that the linearity property also valid for the constant 1 and if it is valid what is the formal calculation of $E(1)$ ?? why is $E(1) = 1$ ???

Can someone also make correction on my $E(Y)$ calculation above because $k$ starts from $1$: $\sum_{k=1}^{\infty}kq^{k-1}p $ , does is make difference from the summation of Geometric expectation which starts at $k=0$ ?

Answer 1

first, if the second summation would start at $0$ you would get the same result: $$ k\cdot q^{k-1}\cdot p = 0, \quad \text{if} \ k=0. $$ Concerning $E[1]=1$. You can think about a random variable $Z$ which always take value $1$, so $P(Z=1)=1$. The name random variable is not so appropriate here, because $Z$ is always equal to one and hence not random, but anyway. From a mathematical perspective a random variable is only a measurable function: $$ Z: \Omega \rightarrow \mathbb R: \omega \mapsto 1. $$ Then clearly $E[Z]=1$, since $Z$ can only take the value $1$. This is what is meant by $E[1]$. Using this definition of $E[1]$ you get the results you need and the linearity holds also in this case.