2 values expected value problem 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.
 A: 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.
$$A = A_1\cdot 0.405\cdot (1-0.405) + A_2\cdot (1-0.405)\cdot 0.405 + A_3\cdot0.405\cdot0.405 + A_4(1-0.405)(1-0.405)$$
A: 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:
$$1000 r^2 -20pr^2-20pr\qquad\text{(dollars)}\ .$$
