# How much saving do you need in the end? zero?

Posted by Yuling Yao on Dec 30, 2022.       Tag: zombie

I came across this book called “Die with zero”. Along with many other yolo ideas, the book prompts the attitude that one must maximize net fulfillment over net worth to the extent of “DIE WITH ZERO”. Indeed, as the book pointed out on the cover

In 1957, Nobel Prize Winner, Franco Modigliani, developed the Life-cycle Hypothesis showing the most optimal way of utilizing your wealth is to end with zero.

I don’t know. I would be moderately shocked if there was indeed an academic effort to show this circular-argument-type result. Seems very trivial.

At the risk of a pedantic tone, I think one common flaw is that people ignore uncertainty. I will show that even in this seemingly trivial problem, uncertainty leads to a surprise.

We need some math here. Suppose one has a fixed total income throughout their life, say $C$ dollars. For simplicity, we would assume this person has no investment nor loan; only spending: they spent $0 \leq Y \leq C$ dollars in his lifetime, such that he has $C-Y$ dollars left in the saving account. In addition, the total amount of necessary costs (food, medical, etc) in the late stage is denoted by $X$ dollars. Naturally, we would wish that the saving could cover such necessary costs, and the final balance sheet $C-Y-X$ is not negative. If $C-Y-X=0$ then a die-with-zero situation occurs.

It seems reasonable to assume that the utility comes from the following two parts:

• whether the necessity is satisfied, that is $1(C-Y-X \geq 0)$. If $C-Y<X$, then the end-of-life bill cannot be paid, bad.
• the spending of leisure: It is a consumerism world, so it is certainly an axiom that the $Y$ dollars spending directly contributed to a utility increment equal to$Y$.

The combined utility is then $1(C-Y-X \geq 0) + Y$, subject to $0 \leq Y \leq C$.

The uncertainty comes from $X$: we do not know the necessary cost at death. Indeed, if X is known, then clearly

$\mathrm{argmax}_{0 \leq Y \leq C} 1(C-Y-X \geq 0) + Y = C-X,$

at which one dies with zero. So yes, we shall all die with zero if we have already calculated our life flawlessly.

In practice, $X$ is a random variable. The decision problem maximizes the expected utility

$\mathrm{max}_{0 \leq Y \leq C} \mathrm{E}_{X} [ 1(C-Y-X \geq 0) + Y ].$

You can check your intuition here: with this uncertainty on $X$, shall we expect more savings? That seems what grandma would say, no? But isn’t the utility a linear function and why would uncertainty matter at all?

We assume $X$ is a normal $(0, \sigma)$ random variable. The expected utility function can be written as $\Pr(X \leq C-Y) + Y$. The derivative of this function with respect to $Y$ is $- \frac{1}{\sigma \sqrt {2\pi}} \exp (-\frac{(C-Y)^2}{2\sigma^2}) + 1$.

1. If $\sigma=0$, then yes, there is no uncertainty, such that dying with zero is optimal.
2. If $\sigma$ is not too big, $\sigma \leq 1/ \sqrt {2\pi}$, then the optimum $\hat y= C-\sqrt{ - 2\sigma^2 \log (\sigma \sqrt {2\pi}) }$. The expected saving at death, is $\sqrt{ - 2\sigma^2 \log (\sigma \sqrt {2\pi}) } >0$, not zero. This is the price one pays for the uncertainty.
3. If $\sigma$ is too big, $\sigma > 1/ \sqrt {2\pi}$. Then the derivative of the objective function is always positive hence the optimal $\hat Y=C$. That is, since the future is just too chaotic, one simply adopts a yolo lifestyle and forgot about the necessity at all. The expected saving at death is either $-X$ or 0 depending on the healthcare system.

The bottom line: die-with-zero is an oversimplification. When there is moderate uncertainty of necessary cost, the optimum would prefer extra non-zero saving at death. When the uncertainty is too big, just yolo.