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Unformatted text preview: ECE 302 Division 2. Homework 6 Solutions. Problem 1. Let X and Y be independent random variables. Random variable X has a discrete uniform distribution over the set { 1 , 2 , 3 } , and Y has a discrete uniform distribution over the set { 1 , 3 } . Let V = X + Y , and W = X − Y . (a) Are V and W independent? Explain without calculations. Solution. No. Independence of V and W would imply that: p W  V ( w  v ) = p W ( w ) , which means the distribution of W cannot depend on the value of V . But in fact, unconditionally, W can have several different experimental values, whereas conditioned on V = 6, both X and Y must be 3 and so W = 0 with probability 1. Thus p W  V ( w  6) negationslash = p W ( w ), and therefore V and W are dependent. (b) Find and plot p V ( v ). Also, determine E [ V ] and var( V ). (c) Find and show in a diagram p V,W ( v,w ). Solution. To visualize this situation, it is helpful to look at the joint PMF of X and Y . Since both X and Y are uniform, and since they are independent, each of the six possible pairs of their experimental values has probability 1/6. In addition to these probabilities, let us tabulate the corresponding experimental values v and w of random variables V and W : y \ x 1 2 3 1 1 / 6 v = 2 w = 0 1 / 6 v = 3 w = 1 1 / 6 v = 4 w = 2 3 1 / 6 v = 4 w = − 2 1 / 6 v = 5 w = − 1 1 / 6 v = 6 w = 0 From this table, it is easy to determine the joint probability mass function of V and W : To get the marginal PMF for V , we need to sum p V,W over w , for each v , i.e., take vertical sums in our picture of the joint PMF of V and W : p V ( v ) = 1 6 , v = 2 , 3 , 5 , 6 , 1 3 , v = 4 , , otherwise. The PMF of V is depicted in figure 2. Since it is symmetric about v = 4, we have: E [ V ] = 4. 1 1/6 1/6 1/6 1/6 1/6 1/6 1 221 2 3 5 6 w v 1 4 Figure 1: A sketch of the joint probability mass function of V and W : there are six equally likely pairs of values, each has probability 1/6. The two circled points comprise the event W > 0 considered in Part (d). 1 2 3 4 5 6 7 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 v p V (v) Figure 2: p V ( v ). Also, var( V ) = E [( V − E [ V ]) 2 ] = E [( V − 4) 2 ] = summationdisplay v ( v − 4) 2 p V ( v ) = 1 6 · [(2 − 4) 2 + (3 − 4) 2 + (5 − 4) 2 + (6 − 4) 2 ] + 1 3 · (4 − 4) 2 = 1 6 (4 + 1 + 1 + 4) = 5 3 . These five conditional probability mass functions are depicted in Figure 3. (d) Find E [ V  W > 0]. Solution. The event W > 0 corresponds to the two circled points in the picture of the joint PMF of V and W . Conditioned on this event, there are two equally likely values that V can assume: 3 and 4. The conditional expectation is therefore 3.5....
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 Spring '08
 Donnely
 Probability theory, discrete uniform distribution

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