2 values expected value problem

Question

A bank account has 1000 dollars and get's a 5% interest daily. Every day the account has $0.405$ probability to lose $20$ dollars.

What's the mean of the amount of money in the acount after 2 days?

I thought about calculating $((1000-20*0.405)*1.05-20*0.405)*1.05$

But I think it might be wrong because the probabiltity changes depending on the previous day. I mean that the probability to lose in both days is $0.405^2$, one day lose and the other no lose would be $0.405*(1-0.405)$ to my understaing, so I would need to take it into acount somehow but don't know how.

thank you.

The 20 $ are lost, when they get lost, before or after applying the $1.0105$ multiplication rate? — dan_fulea, Dec 08 '20 at 12:36
Not sure I see the problem. There are only $4$ scenarios, yes? According to whether the Loss happens ($L$) on a given day or not ($W$), we have $LL$, $LW$, $WL$ or $WW$. Just work out the final cash in each case. — lulu, Dec 08 '20 at 12:44
Well, they don't have the same probability, so no. You have to weight each case by the probability of its occurrence. — lulu, Dec 08 '20 at 12:47
Can mean be returned by a value which isn't a final number but a table in this case? — Mishe Mitasek, Dec 08 '20 at 12:49
@MisheMitasek I haven't touched any mathematics for years, so please make sure my answer is correct before using it — Aven Desta, Dec 08 '20 at 12:54

score 3 · Accepted Answer · answered Dec 08 '20 at 12:49

There are four possibilities.

Loses 20\$ on the first day but not on the second day
Loses 20\$ on the second day but not on the first day
Loses 20\$ on both days
Loses nothing on both day

Calculate the amount in each case, $A_1, A_2, A_3$ and $A_4$. Now to get the mean value, you just multiple each amount with its probability of happening. $$A = A_1\cdot 0.405\cdot (1-0.405) + A_2\cdot (1-0.405)\cdot 0.405 + A_3\cdot0.405\cdot0.405 + A_4(1-0.405)(1-0.405)$$

dan_fulea · Answer 2 · 2020-12-08T13:36:40.993

Two days are a small horizon. The random part is modelled by the binomial tree:

                        * [L] (1000-20)*r - 20 ... and finally ((1000-20)*r -20)*r
                      /
       * [L] 1000-20  - * [K] (1000-20)*r      ... and finally (1000-20)*r*r
      /
1000 *                 * [L] 1000 * r - 20     ... and finally (1000*r -20)*r
      \              / 
       * [K] 1000    - * [K] 1000 * r          ... and finally 1000*r*r

with [L] used for LOSE, and [K] for KEEP. Let $p$ be the probability for losing, i.e. for the [L] ramifications, and $q$ the complementary probability.

Above, the multiplication rate is (EDIT: adjusted after the edit in the OP) $$ r = 1 + 5\% =1.05\ . $$ The mean amount of money after two days is thus: $$ \begin{aligned} M &= p^2\cdot((1000-20)r-20)r +pq\cdot(1000-20)r^2 \\ &\qquad\qquad +pq\cdot(1000r-20)r +q^2\cdot 1000r^2 \\ &=1000r^2-20p^2(r+r^2)-20pq(r+r^2) \\ &=1000r^2-20p(r+r^2)(p+q) \\ &=1000r^2-20p(r+r^2) \ . \end{aligned} $$

Later EDIT:

An other way to think about this is as in the OP. The mean value after two days is by linearity $1000r^2$ plus mean value of the possible loss.

The probability to lose $20\$$ in the first day is $p$, and the corresponding mean is $p\cdot r^2\cdot 20\$$, using the convention that we "first lose, then use the daily rate".
The probability to lose $20\$$ in the second day is $p$ again, and the corresponding mean is $p\cdot r\cdot 20\$$, using the same convention.

So the mean, computed as in the OP comes to the same value: $$1000 r^2 -20pr^2-20pr\qquad\text{(dollars)}\ .$$

Yes, it is wrong, but for an other (minor) reason. The daily multiplication factor is $r=1+1.05%=1.0105$, and using it, your answer is in my notation the same one, $$1000r^2-20pr^2-20pr\ .$$ — dan_fulea, Dec 08 '20 at 13:03
my aplogies the interest is 5%, so I just multiply the amount in the end of the day wether loss or no by 1.05 and then by the probability of ll,,kk,lk,kl right? — Mishe Mitasek, Dec 08 '20 at 13:29
Yes, in this case use the absolute multiplication factor $r=1.05$, everything is all right. — dan_fulea, Dec 08 '20 at 13:36

2 values expected value problem

2 Answers2