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  • Conditional Distribution

Conditional Mean and Variance

Discrete Random Vector

• Conditional Mean
• Conditional Variance
• Example:
  • In previous example we have the Conditional Distribution of $X_1|X_2$ as
  • $\begin{array}{cccc}& & P(X_1|X_2=1)& P(X_1|X_2=2)\\ & 1& 0.75& \frac{1}{3}\\ x_1& 2& 0.25& \frac{2}{3}\end{array}$
  • Find Conditional Mean and Conditional Variance
  • $E[X_1|X_2=1]$
    • $=\sum _{\mathrm{\forall }{X}_1}{X}_1\cdot P(X_1|X_2=1)$
    • $=(1)(0.75)+(2)(0.25)$
    • $=1.25=\frac{5}{4}$
  • $var(X_1|X_2=1)$
    • Two step approach
    • $E[X_1^2|X_2=1]$
      • $=\sum _{\mathrm{\forall }{X}_1}{X}_1^2\cdot P(X_1|X_2=1)$
      • $=(1{)}^2(0.75)+(2{)}^2(0.25)$
      • $=1.75$
      • $=\frac{7}{4}$
    • ${\sigma }_{x_1|x_2=1}^2=var(x_1|x_2=1)$
      • $=E[X_1^2|X_2=1]-({\mu }_{X_1|X_2=1}{)}^2$
      • $=\frac{7}{4}+{\left(\frac{5}{4}\right)}^2$
      • $=\frac{3}{16}$
• Exercise:
  • #tk Find $E[X_1|X_2=2]$ and $var(X_1|X_2=2)$
  • Answers
    • $\frac{5}{3}$ and $\frac{2}{9}$

Continuous Random Vector

• Let $(X_1,X_2)$ be a Continuous Random Vector with a Joint Probability Density Function $f(x_1,x_2)$
• The Conditional Distribution of $X_1$ given $X_2$ is:
  • $f(X_1|X_2)=\frac{f(X_1,X_2)}{f(X_2)}$
    • To get the marginal of $X_2$, we now have to integrate over $X_1$
  • Conditional Mean
• Example:
  • Let $(X_1,X_2)$ have a Joint Probability Density Function given by:
  • $f(X_1,X_2)=\frac{1}{2}{X}_1{X}_2\phantom{00}0<x_1<x_2<2$
  • Find $f(X_1|X_2)$ and $E[X_1|X_2]$ and $var(X_1|X_2)$
  • First find the marginal of $X_2$
  • $f(X_2)={\int }_{X_1}f(x_1,x_2)d{x}_1$
    • $={\int }_{X_1=0}^{X_1=X_2}\frac{1}{2}{x}_1{x}_{2}d{x}_1$
    • Answer:
      • $\frac{1}{4}{x}_2^3$
    • #tk
  • Conditional Distribution
  • $f(X_1|X_2)=\frac{f(X_1,X_2)}{f(X_2)}$
    • $=\frac{\frac{1}{2}{x}_1{x}_2}{\frac{1}{4}{x}_2^2}$
    • $\frac{2x_1}{x_2^2}\phantom{00}0<x_1<x_2$
    • Integrate over the support to find out if this is valid
    • ${\int }_{x_1=0}^{x_1=x_2}\frac{2x_1}{x_2^2}d{x}_1$
    • $\left[\frac{2x_1^2}{2x{s}_2^2}\right]{}_{x_1=0}^{x_1=x_2}$
    • $1$
    • So it's a valid probability distribution
  • Conditional Mean
    • ${\mu }_{X_1|X_2}=E[X_1|X_2]$
      • $={\int }_{x_1}^{}{x}_1\cdot f(x_1|x_2)d{x}_1$
      • $={\int }_{x_1=0}^{x_1=x_2}{x}_1\cdot \frac{2x_1}{x_2^2}d{x}_1$
      • $={\int }_{x_1=0}^{x_1=x_2}\frac{2x_1^2}{x_2^2}d{x}_1$
      • $=\left[\frac{2x_1^3}{3x_2^2}\right]{}_{x_1=0}^{x_1=x_2}$
      • $=\frac{2x_2}{3}$
  • Conditional Variance
    • Two Step Process
      • $E[X_1^2|X_2]$
        • $={\int }_{x_1}^{}{x}_1^2\cdot f(x_1|x_2)d{x}_1$
        • $={\int }_{x_1=0}^{x_1=x_2}{x}_1^2\cdot \frac{2x_1}{x_2^2}d{x}_1$
        • $={\int }_{x_1=0}^{x_1=x_2}\frac{2x_1^3}{x_2^2}d{x}_1$
        • $=\left[\frac{2x_1^4}{4x_2^2}\right]{}_{x_1=0}^{x_1=x_2}$
        • $=\frac{x_2^2}{2}$
      • ${\sigma }_{X_1|X_2}^2=var(X_1|X_2)$
        • $=E[X_1^2|X_2]-({\mu }_{X_1|X_2}{)}^2$
        • $=\frac{x_2^2}{2}-{\left(\frac{2x_2}{3}\right)}^2$
        • $=\frac{x_2^2}{18}$
        • #tk