Probability of selling stock

Question

I have 4 stocks $x_1$, $x_2$, $x_3$, $x_4$ where $x_i$ ~ $N(\$5000,2500)$ What should be the probability of me selling all 4 stocks when

1. At least one stock exceeds \$5022
2. All four exceed \$5022
3. The average of all four exceeds \$5022

For 1., that should just be $1-$ the probability of none exceeds right?

$P = 1 - P(X_i \le 5022) = 1 - \Phi(\frac{5022-5000}{\sqrt{2500}}) = 0.33$

That should be it for 1.? I feel like I'm missing something.

Also, can I get some guidance for 2. and 3. please?

Answer 1

1. You're right that it should be $1-$ none exceeds, however you wrote the probability that one doesn't exceed, so the correct answer should be $1-(P(x_i\leq 5022))^4$ (as we assume that $x_i$ are independet).

2. Similarly, due to independence of $x_i$ we got \begin{equation} P(x_1>5022 \wedge x_2>5022 \wedge x_3>5022 \wedge x_4>5022 ) = \Pi_{i=1}^4 P(x_i>5022) = \Pi_{i=1}^4 (1-P(x_i\leq 5022)) \end{equation}

3. We have to use the property of normal distribution: sum of independent normal variables is a normal variable with mean equal to sum of their means (similarly for variance) \begin{equation} P\bigg(\frac{\sum_{i=1}^4x_i}{4} >5022 \bigg) = P\bigg(\frac{Y}{4}>5022\bigg) \end{equation} where $Y \sim N(4*5022,4*2500)$. Thus $\frac{Y}{4}\sim N(5022,2500/4)$.