If there is a probability of (1-gamma) of you dying everyday, what is your expected lifetime?

Question

Study Question: Verify this fact: if, on every day you wake up, there is a probability of $1 − γ$ that today will be your last day, then your expected lifetime is $1/(1 − γ)$ days.

I thought lifetime of $n$ days means surviving $(n-1)$ days and then die on the last day, so $L_n =\gamma^{(n-1)}(1-\gamma)$, so my expected lifetime would be $$L = (1 - γ) + γ(1 - γ) + γ^2(1-\gamma) + γ^3(1-\gamma) + ... $$

If there are infinite terms, then $$L = (1-\gamma)[1 + \gamma + \gamma^2 + \gamma^3 ...] = (1-\gamma)\frac{1}{1-\gamma} = 1$$ because terms in [] are infinite geometric series.

Judging from the right answer, I know $L_n = \gamma^n$ instead of $\gamma^{(n-1)}(1-\gamma)$, but I don't quite understand why.

Thank you in advance for any help!

Background: Prof. Leslie Kaelbling (MIT) asked this question in her lecture notes regarding Markov decision processes (top of p.73 of the attached link https://openlearninglibrary.mit.edu/assets/courseware/v1/caee9a9ff60ed183e4485869c2e88ac4/asset-v1:MITx+6.036+1T2019+type@asset+block/notes_chapter_Sequential_models.pdf)

Answer 1

The expression $\gamma^{n-1}(1-\gamma)$ is the probability that your lifetime is $n$ days, for $n=1,2,\ldots$. If $L$ stands for lifetime then you've obtained $P(L=n)=\gamma^{n-1}(1-\gamma)$, and your calculation confirms that $\sum_{n=1}^\infty P(L=n)=1$. To compute the expected lifetime, use the formula $$E(L)=\sum_{n=1}^\infty n P(L=n).$$ You will find that this sum equals $1/(1-\gamma)$.

Answer 2

Let $\chi_{n}$ be the random variable that is $1$ if you have survived the days prior to day $n$ (indexing days from $0$) and $0$ otherwise. Clearly $\mathbb{E}\left[\chi_{n}\right]=\gamma^{n}$. Thus your expected lifetime is \begin{align*} \mathbb{E}\left[\sum_{n=0}^{\infty}\chi_{n}\right]\ &=\ \sum_{n=0}^{\infty}\mathbb{E}\left[\chi_{n}\right]\\ &=\ \sum_{n=0}^{\infty} \gamma^{n}\\ &=\ \frac{1}{1-\gamma}, \end{align*} as claimed.