Discrete & Continuous Random Variables - Statistics

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Proof for E[X 1 + X 2 ] = EX 1 + EX 2E[X 1 + X 2 ] = ∑ (x 1 + x 2 ) f (x 1 ,x 2 )=∑∑[x 1 f (x 1 ,x 2 ) + x 2 f (x 1 ,x 2 )](x 1 , x 2 )x1 x2= ∑x1∑x2x 1 f (x 1 ,x 2 ) +∑x2∑x1x 2 f (x 1 ,x 2 )= ∑x1x 1[∑x2f (x 1 ,x 2 )]+∑x2x 2[∑x1f (x 1 ,x 2 )]= ∑x1x 1 f X1 (x 1 ) +∑x2x 2 f X2 (x 2 ) = EX 1 + EX 2Note: we just make use of the distributive law a(b + c) = ab + ac at =and that the order of summation does not change the sum at =.The proof for E[X 1 − X 2 ] = EX 1 − EX 2 proceeds along the same lines.21
µ as Probability Distribution CenterFor µ = EX we haveE[X − µ] = EX − µ = µ − µ = 0i.e., the expected deviation of X from its expectation (or mean) is zero.E[X − µ] = ∑xµ(x − µ) f (x) = 0∑ |x − µ| f (x) = ∑ |x − µ| f (x)x>µx
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Discrete & Continuous Random Variables - Statistics

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