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Additional info for Random variables I
Example text
If z > 0, then the frequency is given by ∞ fZ (z) = −∞ h(x, z − x) dx, where the condition 0 ≤ x ≤ y = z − x is reformulated to 0≤x≤ z . 2 Hence, when z > 0, fZ (z) = so z 2 0 h(x, z − x) dx = z 2 e−(z−x) dx = e−z 0 ⎧ z ⎪ − e(−z), ⎨ exp − 2 fZ (z) = ⎪ ⎩ 0, exp z z − 1 = exp − − exp(−z), 2 2 for z ≥ 0, for z < 0. Alternatively we compute the distribution function of Z by the following double integral, FZ (z) = z 2 x=0 z−x e−y dy y=x dx = z 2 x=0 e−x − ex−z dx z z z − e−z exp − 1 = 1 + e−z − 2 exp − 2 2 2 z 2 1 − exp − .
Y2 ∈ ]0, 1[. ) 1 kY1 (y1 ) = y2 =0 1 −y1 e 2 1 0 y12 − y12 y2 + y1 dy2 1 2 1 1 2 y + y1 = y + y1 e−y1 , 2 1 4 1 2 ∞ 1 ∞ 2 y1 − y12 y2 + y1 e−y1 dy1 k (y1 , y2 ) dy1 = 2 y1 =0 0 1 3 (2 − 2y2 + 1) = − y2 . 2 2 1 −y1 e 2 = kY2 (y2 ) k (y1 , y2 ) dy2 = = = y12 − Since k (y1 , y2 ) = kY1 (y1 ) · kY2 (y2 ), we see that Y1 and Y2 are not independent. 5) The means are E {Y1 } = E {X1 + X2 } = E {X1 } + E {X2 } = 5 3 +1= . 2 2 Alternatively, E {Y1 } = ∞ ∞ y1 kY1 (y1 ) dy1 = 0 0 5 6 2 1 3 1 2 −y1 y + y e dy1 = + = .
X2 1 . X1 1. Compute the means E {X1 } and E 2. Compute the means of Y1 and for Y2 . The vector function τ , given by τ (x1 , x2 ) = x1 x2 , x1 x2 , maps D = ]0, 2[ × ]0, 2[ bijectively onto D = (y1 , y2 ) ∈ R2 | 0 < y1 < 4y2 , y1 y2 < 4 . 3. Sketch D , and find the simultaneous frequency k (y1 , y2 ) of (Y1 , Y2 ). 4. Find the marginal frequencies of Y1 and Y2 . 5. Are Y1 and Y2 independent? 1) It follows that E {X1 } = 2 1 2 x2 dx = 0 1 2 x3 3 2 = 0 4 , 3 and E 1 X1 = 1 2 2 0 x dx = 1. x 2) Since X1 and X2 are independent, we conclude from (1) that 4 3 E {Y1 } = E {X1 X2 } = E {E1 } E {X2 } = 3 = 16 , 9 and E {Y2 } = E X1 X2 = E {X1 } · E 1 X2 = 4 4 ·1= .