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