A man bought $4$ stocks, the value of the first stock is $100\$ $ , the second $150\$$, third $50\$$ and the fourth $20\$$.
We define $v_i$ to be the value of the stock $i$.
The probability for the value of the stock rises in the end of the week is $\frac{\sqrt{v_i}}{60}$ independently from all the other stocks.
Let $X$ be the number of stocks the man has won money in the end of the week.
Calculate:
-The probability that the man has won in at least $2$ stocks.
-The expected value of $X$, and find an upper bound of the probability you found above using Markov's inequality.
-The variance of $X$, and find an upper bound of the probability you found above using Chebyshev inequality.
My Work:
Calculating $P(X\ge 2)$:
$P(X\ge 2) = P(X=2)+P(X=3)+P(X=4) = \frac{\sqrt{100}}{60}\frac{\sqrt{150}}{60}(1-\frac{\sqrt{50}}{60})(1-\frac{\sqrt{20}}{60}) + \frac{\sqrt{150}}{60}\frac{\sqrt{50}}{60}(1-\frac{\sqrt{100}}{60})(1-\frac{\sqrt{20}}{60})+ \dots (\text{All the possibilities of choosing $2$ out of the $4$)} + \frac{\sqrt{100}}{60}\frac{\sqrt{150}}{60}\frac{\sqrt{50}}{60}(1-\frac{\sqrt{20}}{60})+ \dots (\text{All the possibilities of choosing 3 out of the 4)} +\frac{\sqrt{100}}{60}\frac{\sqrt{150}}{60}\frac{\sqrt{50}}{60}\frac{\sqrt{20}}{60}$
I'm pretty unsure that this is the most efficient solution or if it's the right one, would appreciate any feedback.
For the next steps,
I thought of defining $I_1,I_2,I_3,I_4$ as indicators of stocks $1$ to $4$, in which each indicator is equal to one if the man has won in the stock and zero if not.
So $X= \sum^4_{i=1}I_i$ .
Expected value calculation:
Then I can say that $E(X)=E(\sum^4_{i=1}I_i)=\sum^4_{i=1}E(I_i)=\sum^4_{i=1}P(I_i=1)=\sum^4_{i=1} \frac{\sqrt{v_i}}{60}$
Variance Calculation:
$var(X) = E(X^2)-(E(X))^2$ So I need to calculate $E(X^2)$
$E(X^2)=E(\sum^4_{i=1}I_i\sum^4_{j=1}I_j) = E(\sum^4_{i=1}I_i^2 + 2 \sum_{i<j}I_iI_j)=E(\sum^4_{i=1}I_i^2) + 2E( \sum_{i<j}I_iI_j)=E(X)+ 2*\sum_{i<j}P(I_i=1)P(I_j=1)=E(X)+2 \sum_{i<j}\frac{\sqrt{v_i}}{60} \frac{\sqrt{v_j}}{60}$
Inequalities:
I haven't yet tried to calculate that long probability of $P(X\ge 2)$, because I'm pretty unsure of it, and so I still haven't tried to find upper bounds using the inequalities. But I've tried to set the inequalities and finding some trouble with Chebyshev's inequality:
$P(X\ge 2) \le \frac{E(X)}{2}$ -- Markov's inequality.
$P(|X-E(X)|\ge a) \le\frac{var(X)}{a^2}$ for any $a>0$, I'm not sure here how can I reach $P(X\ge 2)$.
I am really unsure of this long solution, and would appreciate any approval / dissapproval of any step I made or approach, so I can learn from it and turn it into a better solution.
Thanks in advance.
On a quick read through, your calculations look good to me, and I think using indicator functions, as you've done, is probably the best way to go about calculating the mean and variance of $\ X\ $. You can simplify your calculation of the variance, however. Since $\ I_1, I_2, I_3, I_4\ $ are independent, the variance of their sum is the sum of their variances, so you can avoid having to evaluate your more complicated expression for $\ E\big(X^2\big)\ $