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if discrete random variable X is uniformly distributed over {-7,-5,-3,-1,1,3,5,7},then how to calculate the expectation of X and mod(X) and also expectation of X^2 and mod(X^2).It would be appreciable if it is diagramatically explained.

Ram Vanka
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  • What have you tried? This should easily follow from the definitions of uniform distribution and expectation. – M Turgeon Aug 19 '13 at 20:25
  • Mr.Turegon I am having some problems in calculating the mean and second order moment,otherwise I wouldn't have approached this website.Please help,I basically need answer about the case of a RV,Z=X+jY,where j=sqrt(-1) and X and Y are independent and both are uniformly distributed over {-7,-5,-3,-1,1,3,5,7}.Please help me with E(Z),E(abs(Z)),E((abs(Z))^2).I am in very much need of the calculations involved. – Ram Vanka Aug 21 '13 at 04:08
  • If what you need is the answer, this is not the right website for you. If you have trouble understanding the solution below, ask for clarifications. But I feel like the main problem here is with your understanding of the definitions. – M Turgeon Aug 21 '13 at 14:12

1 Answers1

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The expected value of X is pretty straightforward. First off, we know that X is discrete, and it takes values of:

-7, -5, -3, -1, 1, 3, 5, 7

Uniform distribution of a discrete random variable means that each of these values of X is equally likely to occur. Since there are 8 possible values of X, the probability that X takes on any one of the 8 values is: 1/8.

Following this logic, the expected value of X is simply:

$$ EX = (-7)*(1/8) + (-5)*(1/8) + ...+(7)*(1/8) $$

Ironically, since the range of values of X straddles zero, this evaluation will give you that EX = 0.

Now, for E(X^2), you might be tempted to take EX and square it. Resist the urge, and instead take all the possible values of X and square them. As a result of the fact that y^2 and (-y)^2 equate to the same number, there are now only 4 possible values of X^2:

$$ (-7)^2 = (7)^2 = 49, (-5)^2 = (5)^2 = 25, (-3)^2 = (3)^2 = 9, (-1)^2 = (1)^2 = 1 $$

Again, go through the same process that we did above to find EX, but in this case, there are only 4 possible values, so the probability of each one is only 1/4, not 1/8 as before. Once you've done that, you will have your answer for E(X^2).

Now, for mod(x), I'm not quite sure what you mean by that. Please leave a comment or something to clarify; hope this has helped.

user79790
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  • Probably $\text{mod}(X)$ is intended to mean $|X|$. – André Nicolas Aug 19 '13 at 20:32
  • If @AndréNicolas is correct, then the only change you need to make is to take the absolute value of each possible value of X, leaving the probabilities the same, and then evaluate the sum for EX. Again, all of this follows directly from the definition of Uniform Distribution and the definition of Expectation, with respect to Discrete Random Variables. The process changes slightly when dealing with Continuous Random Variables in that integration is used. – user79790 Aug 19 '13 at 20:39
  • Could also be the mode of $X$... – gt6989b Aug 19 '13 at 20:43
  • Anything is possible, but since every value is a mode in this case, highly unlikely! – André Nicolas Aug 19 '13 at 21:53
  • Yes,I meant abs(X) by "mod(X)".Can you please explain the calculation of E(abs(X)) and also E((abs(X))^2).Also please tell me what would be the case of a RV,Z=X+jY,where j=sqrt(-1) and X and Y are independent and both are uniformly distributed over {-7,-5,-3,-1,1,3,5,7}.Please help me with E(Z),E(abs(Z)),E((abs(Z))^2).I am in very much need of these answers. – Ram Vanka Aug 20 '13 at 07:39