STA256 Tutorial W10

2.8.7
- RV $X, Y$
- with $μ_{1} = 1$ and $μ_{2} = 4$
- with $σ_{1}^{2} = 4$ and $σ_{2}^{2} = 6$ , $ρ = \frac{1}{2}$
- $Z = 3 X - 2 Y$
- Find $E [Z]$
  - $E [Z] = μ_{3}$
  - Transformation
    - $Z = 3 X - 2 Y$
  - Inverse
    - $Z - 2 Y = 3 X$
    - $U = X$
  - $E [Z] = E [3 X - 2 Y]$
  - $E [Z] = E [3 X] - E [2 Y]$
  - $E [Z] = 3 E [X] - 2 E [Y]$
  - $E [Z] = 3 - 2 (4) = - 5$
- Find $v a r (Z)$
  - $v a r (Z) = σ^{2} = E [Z^{2}] - E [Z]^{2}$
  - $v a r (Z) = σ^{2} = E [(3 X - 2 Y)^{2}] - (- 5)^{2}$
  - $v a r (Z) = σ^{2} = E [9 X^{2} - 12 X Y + 4 Y^{2}] - (- 5)^{2}$
  - $v a r (Z) = v a r (3 X - 2 Y)$
    - $S = \sum_{i = 1}^{n} a x$
    - $v a r (S) = \sum_{i = 1}^{n} a_{i}^{2} v a r (X_{i})$
      - $X ⊥ Y ⟹ c o v (x, y) = 0$
    - $v a r (S) = \sum_{i = 1}^{n} a_{i}^{2} v a r (X_{i}) + 2 \sum_{i = 1}^{n} \sum_{j = 0}^{n} a_{i} a_{j} c o v (x_{i}, x_{j})$
  - $v a r (Z) = v a r (3 X - 2 Y)$
    - $= 3^{2} v a r (x) + 4 v a r (y) + 2 (3) (- 2) c o v (x, y)$
      - $ρ = \frac{c o v (x_{1}, x_{2})}{σ_{x_{1}} σ_{x_{2}}}$
      - $ρ = \frac{c o v (x, y)}{σ_{1} σ_{2}}$
      - $ρ = \frac{c o v (x, y)}{2 \sqrt{6}}$
      - $\frac{1}{2} = \frac{c o v (x, y)}{2 \sqrt{6}}$
      - $\frac{2 \sqrt{6}}{2} = c o v (x, y)$
      - $\sqrt{6} = c o v (x, y)$
    - $= 9 (4) + 4 (6) - 12 (\sqrt{6})$
    - $= 60 - 12 (\sqrt{6})$
2.8.11
2.6.1
- $X_{1}, X_{2}, X_{3}$ RVs
- $M_{X} (t) = e^{2 t + \frac{t^{2}}{2}}$ for $t \in R$
- Find $M_{T} (t)$ where $T = 2 X_{1} - X_{2} + X_{3}$
- Theorem
  - $X_{1}, \dots, X_{n}$ are independent
  - $X_{i}$ has mgf
  - $M_{T} (t) = \prod_{i = 1}^{n} M_{X} (k_{i} t)$
- $M_{T} (t) = M_{X} (2 t) M_{X} (- t) M_{X} (t)$
- $M_{T} (t) = (e^{4 t + \frac{4 t^{2}}{2}}) (e^{- 2 t + \frac{t^{2}}{2}}) (e^{2 t + \frac{t^{2}}{2}})$
- $M_{T} (t) = e^{(4 t + \frac{4 t^{2}}{2}) + (- 2 t + \frac{t^{2}}{2}) + (2 t + \frac{t^{2}}{2})}$
- $M_{T} (t) = e^{4 t + \frac{4 t^{2} + t^{2} + t^{2}}{2}}$
- $M_{T} (t) = e^{4 t + \frac{6 t^{2}}{2}}$
- $M_{T} (t) = e^{4 t + 3 t^{2}}$
- $M_{T} (t) = e^{t (4 + 3 t)}$
3.1.6
- Let $Y$ be the number of successes throughout $n$ independent repetitions of experiment where probability $p = \frac{1}{4}$
  - $x_{1}, \dots, x_{n}$
- Determine the smallest $n$ such that $P (Y \geq 1) \geq 0.7$
- $B (n, p)$
- $P (X = x) = (\binom{n}{x}) p^{x} (1 - p)^{n - x}$
- $P (Y = y) = (\binom{n}{x}) {0.25}^{x} (0.75)^{n - x}$
- $P (Y \geq 1) = 1 - P (Y = 0)$
- $P (Y \geq 1) = P (Y = 1) = P (X = 1) = \frac{1}{4}$ ?
- $P (Y \geq 1) = 1 - P (Y = 0)$
  - We went through $n$ repetitions and all of the fail
  - $P (Y = 0) = (\binom{n}{0}) {0.25}^{y} (0.75)^{n - y}$
  - $1 - P (Y = 0)$
  - $1 - {(\frac{3}{4})}^{n}$
    - Failure rate to the $n$
  - $1 - {(\frac{3}{4})}^{n} \geq 0.7$
  - $1 - {(\frac{3}{4})}^{n} \geq 0.7$
  - $- {(\frac{3}{4})}^{n} \geq - 0.3$
  - ${(\frac{3}{4})}^{n} \leq 0.3$
  - Lowest value possible is $n = 5$
  - Calculate by log of both sides
  - $n \log \frac{3}{4} = \log 0.3$
  - $n = \frac{\log 0.3}{\log \frac{3}{4}} \approx 4.185$
  - $n \geq 4.185$
  - Since it's discrete
  - Round up