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Y=mX+cY = mX + cY=mX+c
E(kX)=kE(X)E(kX) = kE(X)E(kX)=kE(X)
Var(kX)=k2Var(X)\text{Var}(kX) = k^2 \text{Var}(X)Var(kX)=k2Var(X)
E(X+k)=E(X)+kE(X + k) = E(X) + kE(X+k)=E(X)+k
Var(X+k)=Var(X)\text{Var}(X + k) = \text{Var}(X)Var(X+k)=Var(X)
(X+Y)∼N(μx+μy,σx2+σy2)(X + Y) \sim N(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)(X+Y)∼N(μx+μy,σx2+σy2)
(X−Y)∼N(μx−μy,σx2+σy2)(X - Y) \sim N(\mu_x - \mu_y, \sigma_x^2 + \sigma_y^2)(X−Y)∼N(μx−μy,σx2+σy2)
(X+Y)∼Po(λx+λy)(X + Y) \sim Po(\lambda_x + \lambda_y)(X+Y)∼Po(λx+λy)
Mean equals variance
No, because E(X−Y)≠Var(X−Y)E(X - Y) \neq \text{Var}(X - Y)E(X−Y)=Var(X−Y)
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