text: # Conditional Expectation Help
  
  • Dec 8th 2010, 04:59 PM
  theprestige
  Conditional Expectation Help
  Suppose you flip one fair coin and roll one fair six-sided die. Let X be the number showing on the die, and let Y = 1 if the coin is heads with Y = 0 if the coin is tails. Let Z = XY.
  
  Compute E(Z):
  
  Am I correct when I do E(Z) = E(X)E(Y) = (3.5)(1/2) = 7/4
  
  Also, how do i do this?
  
  Compute E(Z|X = 4)
  
  Thanks
  • Dec 9th 2010, 04:52 AM
  Focus
  Quote:
  
  Originally Posted by theprestige
  Suppose you flip one fair coin and roll one fair six-sided die. Let X be the number showing on the die, and let Y = 1 if the coin is heads with Y = 0 if the coin is tails. Let Z = XY.
  
  Compute E(Z):
  
  Am I correct when I do E(Z) = E(X)E(Y) = (3.5)(1/2) = 7/4
  
  I am guessing X and Y are independent, in which case you are correct.
  
  Quote:
  
  Also, how do i do this?
  
  Compute E(Z|X = 4)
  
  Thanks
  Given that you rolled a 4, what is the expectation of Z? There are only 2 values Y could take with equal probability, so look at what happens at each. You could look at this as $\mathbb{E}[4Y]$.
metadata: {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}