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differentiating with respect to θthe expressions on both sides of the equation from x=1 to ∞Σθ(1-θ)^x-1=1 show that the mean of the geometric distribution is given by μ=1/θthen differentiating again with respect to θ show that μ2 ' =2-θ/θ^2 and hence that σ^2=1-θ/θ^2
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[x=1 to ∞] Σθ(1-θ)^x-1=1 對θ微分[x=1 to ∞] Σ[θ(x-1)(1-θ)^x-2（-1）+(1-θ)^x-1]=0…(1)[x=1 to ∞] Σ(1-θ)^x-2 [θ(1-x)+(1-θ)]=0[x=1 to ∞] Σ(1-θ)^x-2 (1-x)=0 等號兩邊同乘θ（1-θ）[x=1 to ∞] Σθ(1-θ)^x-1 (1-x)=0[x=1 to ∞] Σθ(1-θ)^x-1 - [x=1 to ∞] Σxθ(1-θ)^x-1=01/[1-(1-θ)]-E（x）=0E(x)=1/θ (1) 對θ微分[x=1 to ∞] Σ[θ(1-x)(2-x)(1-θ)^x-3+(1-θ)^x-2+(1-x) (1-θ)^x-2]=0[x=1 to ∞] Σ(1-x) (1-θ)^x-2 [θ (2-x)(1-θ)^-1+1+1]=0等號兩邊同乘（1-θ）[x=1 to ∞] Σ(1-x) (1-θ)^x-1 [θ (2-x) +2(1-θ）]=0[x=1 to ∞] Σ(1-x) (1-θ)^x-1 (2-θx)=0[x=1 to ∞] Σ(2+θx^2-2x-θx) (1-θ)^x-1 =0等號兩邊同乘θ後展開[x=1 to ∞] ｛2Σθ (1-θ)^x-1 +θΣx^2θ (1-θ)^x-1-2Σxθ (1-θ)^x-1-θΣxθ (1-θ)^x-1=02+θE（x^2）-2E(x)- θE（x）=0-θE（x^2）=2-2*1/θ-θ*1/θ=1-2/θ=(θ-2)/ θμ2 '=E（x^2）=(2-θ)/ θ^2σ^2=μ2 '-μ1 '^2=(2-θ)/θ^2-(1/θ)^2=(1-θ)/θ^2