The Normal and LognormalDistributionsJohn [email protected]://www.norstad.orgFebruary 2, 1999Updated: November 3, 2011AbstractThe basic properties of the normal and lognormal distributions, with full proofs.We assume familiarity with elementary probability theory and with college-levelcalculus.1 DEFINITIONS AND SUMMARY OF THE PROPOSITIONS 11 Definitions and Summary of the PropositionsProposition 1:Z∞−∞1√2πσe−(x−µ)2/2σ2dx = 1Proposition 2:Z∞−∞x1√2πσe−(x−µ)2/2σ2dx = µProposition 3:Z∞−∞x21√2πσe−(x−µ)2/2σ2dx = µ2+ σ2Definition 1 The normal distribution N[µ, σ2] is the probability distributiondefined by the following density function:1√2πσe−(x−µ)2/2σ2Note that Proposition 1 verifies that this is a valid density function (its integralfrom −∞ to ∞ is 1).Definition 2 The lognormal distribution LN [µ, σ2] is the distribution of eXwhere X is N[µ, σ2].Proposition 4: If X is N[µ, σ2] then E(X) = µ and Var(X) = σ2.Proposition 5: If Y is LN[µ, σ2] then E(Y ) = eµ+12σ2andVar(Y ) = e2µ+σ2(eσ2− 1).Proposition 6: If X is N[µ, σ2] then aX + b is N [aµ + b, a2σ2].Proposition 7: If X is N[µ1, σ21], Y is N[µ2, σ22], and X and Y are indepen-dent, then X + Y is N[µ1+ µ2, σ21+ σ22].Corollary 1: If Xiare independent N[µ, σ2] for i = 1 . . . n thennXi=1XiisN[nµ, nσ2].Corollary 2: If Yiare independent LN [µ, σ2] for i = 1 . . . n thennYi=1YiisLN[nµ, nσ2].Proposition 8: The probability density function of LN[µ, σ2] is:1x√2πσe−(log(x)−µ)2/2σ22 PROOFS OF THE PROPOSITIONS 22 Proofs of the PropositionsProposition 1Z∞−∞1√2πσe−(x−µ)2/2σ2dx = 1Proof:First assume that µ = 0 and σ = 1. Let:a =Z∞−∞1√2πe−x2/2dxThen:a2=Z∞−∞Z∞−∞1√2πe−x2/21√2πe−y2/2dxdy=12πZ∞−∞Z∞−∞e−(x2+y2)/2dxdyApply the polar transformation x = r cos θ, y = r sin θ, dxdy = rdrdθ:a2=12πZ2π0Z∞0e−r2/2rdrdθ=12πZ2π0h−e−r2/2i∞0dθ=12πZ2π0[0 − (−1)]dθ=12π2π= 1a > 0, and we just showed that a2= 1, so we must have a = 1.For the general case, apply the transformation y =x − µσ, dy =dxσ:Z∞−∞1√2πσe−(x−µ)2/2σ2dx =Z∞−∞1√2πσe−y2/2σdy=Z∞−∞1√2πe−y2/2dy= 12 PROOFS OF THE PROPOSITIONS 3Proposition 2Z∞−∞x1√2πσe−(x−µ)2/2σ2dx = µProof:First assume that µ = 0 and σ = 1.Z∞−∞x1√2πe−x2/2dx = limn→∞1√2πZn−nxe−x2/2dx=1√2πlimn→∞h−e−x2/2in−n=1√2πlimn→∞[(−e−n2/2) − (−e−n2/2)]=1√2πlimn→∞0= 0For the general case, apply the transformation y =x − µσ, dy =dxσ:Z∞−∞x1√2πσe−(x−µ)2/2σ2dx =Z∞−∞(µ + σy)1√2πσe−y2/2σdy= µZ∞−∞1√2πe−y2/2dy + σZ∞−∞y1√2πe−y2/2dy= µ × 1 + σ × 0 (by Proposition 1)= µProposition 3Z∞−∞x21√2πσe−(x−µ)2/2σ2dx = µ2+ σ2Proof:First assume that µ = 0 and σ = 1.Integrate by parts using f = x, f0= 1, g = −e−x2/2, g0= xe−x2/2:Zn−nx2e−x2/2dx =Zn−nfg0dx= f(n)g(n) − f(−n)g(−n) −Zn−nf0gdx= (−ne−n2/2) − (ne−n2/2) −Zn−n−e−x2/2dx2 PROOFS OF THE PROPOSITIONS 4= −2ne−n2/2−Zn−n−e−x2/2dx=Zn−ne−x2/2dx − 2ne−n2/2Then:Z∞−∞x21√2πe−x2/2dx =1√2πlimn→∞Zn−nx2e−x2/2dx=1√2πlimn→∞Zn−ne−x2/2dx − 2ne−n2/2=Z∞−∞1√2πe−x2/2dx − limn→∞2ne−n2/2= 1 − limn→∞2ne−n2/2(by Proposition 1)All that remains is to show that the last limit above is 0. We do this usingL’Hˆopital’s rule:limn→∞2ne−n2/2= limn→∞2nen2/2= limn→∞2nen2/2= 0For the general case, apply the transformation y =x − µσ, dy =dxσ:Z∞−∞x21√2πσe−(x−µ)2/2σ2dx =Z∞−∞(µ + σy)21√2πσe−y2/2σdy= µ2Z∞−∞1√2πe−y2/2dy +2µσZ∞−∞y1√2πe−y2/2dy +σ2Z∞−∞y21√2πe−y2/2dy= µ2× 1 + 2µσ × 0 + σ2× 1(by Propositions 1 and 2)= µ2+ σ22 PROOFS OF THE PROPOSITIONS 5Proposition 4 If X is N[µ, σ2] then E(X) = µ and Var(X) = σ2.Proof:E(X) =Z∞−∞x1√2πσe−(x−µ)2/2σ2dx = µ (by Proposition 2)Var(X) = E(X2) − E(X)2(by Proposition 1 in reference [1])=Z∞−∞x21√2πσe−(x−µ)2/2σ2dx − µ2= µ2+ σ2− µ2(by Proposition 3)= σ2Proposition 5 If Y is LN[µ, σ2] then E(Y ) = eµ+12σ2andVar(Y ) = e2µ+σ2(eσ2− 1).Proof:Y = eXwhere X is N [µ, σ2]. First assume that µ = 0:E(Y ) = E(eX) =Z∞−∞ex1√2πσe−x2/2σ2dx=Z∞−∞1√2πσe2xσ2−x22σ2dx=Z∞−∞1√2πσe−(x−σ2)2+σ42σ2dx= e12σ2Z∞−∞1√2πσe−(x−σ2)2/2σ2dx= e12σ2× 1 (by Proposition 1)= e12σ2E(Y2) = E(e2X) =Z∞−∞e2x1√2πσe−x2/2σ2dx=Z∞−∞1√2πσe4xσ2−x22σ2dx=Z∞−∞1√2πσe−(x−2σ2)2+4σ42σ2dx= e2σ2Z∞−∞1√2πσe−(x−2σ2)2/2σ2dx= e2σ2× 1 (by Proposition 1)= e2σ2Var(Y ) = E(Y2) − E(Y )2(by Proposition 1 in reference [1])= e2σ2− (e12σ2)2= e2σ2− eσ2= eσ2(eσ2− 1)2 PROOFS OF THE PROPOSITIONS 6For the general case, apply the transformation y = x − µ, dy = dx:E(Y ) = E(eX) =Z∞−∞ex1√2πσe−(x−µ)2/2σ2dx=Z∞−∞eµ+y1√2πσe−y2/2σ2dy= eµZ∞−∞ey1√2πσe−y2/2σ2dy= (by the same reasoning as for the case µ = 0 above)eµe12σ2= eµ+12σ2E(Y2) = E(e2X) =Z∞−∞e2x1√2πσe−(x−µ)2/2σ2dx=Z∞−∞e2(y+µ)1√2πσe−y2/2σ2dy= e2µZ∞−∞e2y1√2πσe−y2/2σ2dy= (by the same reasoning as for the case µ = 0 above)e2µe2σ2= e2µ+2σ2Var(Y ) = E(Y2) − E(Y )2(by Proposition 1 in reference [1])= e2µ+2σ2− (eµ+12σ2)2= e2µ+2σ2− e2µ+σ2= e2µ+σ2(eσ2− 1)Proposition 6 If X is N[µ, σ2] then aX + b is N[aµ + b, a2σ2].Proof:Prob(aX + b < k) = Prob(X < (k − b)/a)=Z(k−b)/a−∞1√2πσe−(x−µ)2/2σ2dx= (apply the transformation y = ax + b, dy = adx)Zk−∞1√2πσe−(y−ba−µ)2/2σ21ady=Zk−∞1√2π(aσ)e−(y−(aµ+b))2/2a2σ2dyThe last term above is the cumulative density function for N[aµ + b, a2σ2], sowe have our result.2 PROOFS OF THE PROPOSITIONS 7Proposition 7 If X is N[µ1, σ21], Y is N[µ2, σ22], and X and Y are indepen-dent, then X + Y is N [µ1+ µ2, σ21+ σ22].Proof:First assume that X is N [0, 1] and Y is N[0, σ2]. Then:Prob(X + Y < k) =Z Zx+u<k1√2πe−x2/21√2πσe−u2/2σ2dxdu= (apply the transformation u = σy)12πσZ Zx+σy<ke−(x2+y2)/2σdxdy=12πZ Zx+σy<ke−(x2+y2)/2dxdy (1)At this point we temporarily make the assumption that k ≥ 0. Figure 1 showsthe area