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  • STA256 Lecture 15

Independent Random Vectors

• Independent if:
  • $P(A\cap B)=P(A)P(B)$

Discrete Random Vector

• Let $(X_1,X_2)$ be a Discrete Random Vector, with a Joint Probability Mass Function of $P(X_1,X_2)$.
• $X_1$ and $X_2$ are independent if:
  • $P(X_1=x_1,X_2=x_2)=P(X_1=x_1)P(X_2=x_2)$
  • for the entire support of $X_1$ and $X_2$
• Example:
  • Let $(X_1,X_2)$ have a Joint Probability Mass Function
  • $\begin{array}{ccc}& & X_2\\ & & 3& 4\\ X_1& 1& 0.2& 0.2& 0.4\\ & 2& 0.3& 0.3& 0.6\\ & & 0.5& 0.5\end{array}$
  • #tk The Marginal Distributions are:
    • $\begin{array}{ccc}{X}_1=x_1& 1& 2\\ P(X_1=x_1)& 0.4& 0.6\end{array}$
    • $\begin{array}{ccc}{X}_2=x_2& 3& 4\\ P(X_2=x_2)& 0.5& 0.5\end{array}$
  • Determine whether $X_1$ and $X_2$ are independent
  • $\begin{array}{cc}\text{Product of Marginals}& \text{Joint}\\ P(X_1=x_1)P(X_2=x_2)& P(X_1=x_1,X_2=x_2)\\ \underset{―}{P(X_1=1)P(X_2=3)}& \underset{―}{P(X_1=1,X_2=3)}\\ (0.4)(0.5)=0.2& 0.2& ✓\\ \underset{―}{P(X_1=1)P(X_2=4)}& \underset{―}{P(X_1=1,X_2=4)}\\ (0.4)(0.5)=0.2& 0.2& ✓\\ \underset{―}{P(X_1=2)P(X_2=3)}& \underset{―}{P(X_1=2,X_2=3)}\\ (0.6)(0.5)=0.3& 0.3& ✓\\ \underset{―}{P(X_1=2)P(X_2=4)}& \underset{―}{P(X_1=2,X_2=4)}\\ (0.6)(0.5)=0.3& 0.3& ✓\end{array}$
  • They're equal over the entire support, so they're independent.
• Exercise:
  • Example from 2.3
  • Show that $X_1$ and $X_2$ are not independent.
  • #tk

Continuous Random Vector

• Let $(X_1,X_2)$ be a Continuous Random Vector, with a Joint Probability Density Function of $f(X_1,X_2)$.
• To show that $X_1$ and $X_2$ are independent
• $f(x_1,x_2)=f(x_1)f(x_2)$
• Example:
  • Let $(X_1,X_2)$ be Continuous Random Vector with a Joint Probability Density Function given by:
  • $f(X_1,X_2)=\frac{X_1{X}_2^2}{18},\begin{array}{c}0\le x_1\le 2\\ 0\le x_2\le 3\end{array}$
  • Determine whether they're independent
  • Find the Marginal Distributions
    • $f(x_1)={\int }_{x_2}^{}f(x_1,x_2)d{x}_2$
      • $={\int }_{x_2=0}^{x_2=3}\frac{x_1{x}_2^2}{18}d{x}_2$
      • $=\left[\frac{x_1{x}_2^2}{54}\right]{}_{x_2=0}^{x_2=3}$
      • $\frac{x_1}{2},0\le x_1\le 2$
    • $f(x_2)={\int }_{x_1}^{}f(x_1,x_2)d{x}_1$
      • $={\int }_{x_1=0}^{x_1=2}\frac{x_1{x}_2^2}{18}d{x}_1$
      • $=[\dots ]$
        • $=\frac{x_2^2}{9},0\le x_2\le 3$
  • Verify if they're the same or not.
  • $\begin{array}{cc}\text{Product of Marginals}& \text{Joint}\\ f(x_1)f(x_2)& f(x_1,x_2)\\ \underset{―}{\left(\frac{x_1}{2}\right)\left(\frac{x_2^2}{9}\right)}& \underset{―}{\frac{x_1{s}_2^2}{18}}\\ \frac{x_1{x}_2^2}{18}& \frac{x_1{x}_2^2}{18}& ✓\end{array}$
  • Since they're equal, they're independent.
  • #tk show from example in 2.3 they're not independent.
    • You can also see that in the support the support of one depends on the support of the other.

Theorem 19: Joint Cumulative Distribution Function

• #tk Theorem 19 is another way to show independence
• Let $(X_1,X_2)$ be Continuous Random Vector with a Joint Probability Density Function given by:
• $f(X_1,X_2)=\frac{X_1{X}_2^2}{18},\begin{array}{c}0\le x_1\le 2\\ 0\le x_2\le 3\end{array}$
• Joint Cumulative Distribution Function
• $F(X_1,X_2)={\int }_{t_2=?}^{t_2=x_2}{\int }_{t_1=?}^{t_1=x_1}f(t_1,t_2)d{t}_{1}d{t}_2$
  • $={\int }_{t_2=0}^{t_2=x_2}{\int }_{t_1=0}^{t_1=x_1}\frac{t_1{t}_2^2}{18}d{t}_{1}d{t}_2$
  • $={\int }_{t_2=0}^{t_2=x_2}\left[\frac{t_1^2{t}_2^2}{36}\right]{}_{t_1=0}^{t_1=x_1}d{t}_2$
  • $={\int }_{t_2=0}^{t_2=x_2}\frac{x_1^2{t}_2^2}{36}d{t}_2$
  • $\dots$
  • #tk
  • $=\frac{x_1^2{x}_2^3}{108}$
• Marginal Cumulative Distribution Function
• From earlier work, we obtained $f(x_1)=\frac{x_1^2}{2},0\le x_1\le 2$
• $f(x_2)=\frac{x_2^2}{9},0\le x_2\le 3$
• $F(x_1)={\int }_{t_1=-\mathrm{\infty }}^{t_1=x_1}d{t}_1$
  • $={\int }_{t_1=0}^{t_1=x_1}\frac{t_1^2}{2}d{t}_1$
  • $=\frac{x_1^2}{4}$
• We can show that $F(x_2)=\frac{x_2^3}{27}$
• Applying theorem 19
• $\begin{array}{cc}\text{Product of Marginal CDFs}& \text{Joint of Marginal CDFs}\\ \underset{―}{F(X_1)F(X_2)}& \underset{―}{F(X_1,X_2)}\\ \left(\frac{X_1^2}{4}\right)\left(\frac{X_2^3}{27}\right)=\frac{x_1^2{x}_2^3}{108}& \frac{x_1^2{x}_2^3}{108}& ✓\end{array}$