STAT 400

Mon. February 17th, 2020


Variance and Standard Deviation

We use the concept of variance to measure of the deviation of a random variable XX from its expected value μ\mu.

Definition. Let XX be a discrete random variable with probability mass function p(x)p(x) and expected value μ\mu. The variance of XX, denoted by V(X)V(X) or sometimes Var(X)\text{Var}(X) (or σX2\sigma_X^2 or σ2\sigma^2) is V(X)=∑y∈D(y−μ)2⋅p(y)=E((x−μ)2).V(X)=\sum_{y\in D}(y-\mu)^2\cdot p(y)=E((x-\mu)^2).
We define the standard deviation (SD) of XX as σX=σX2=V(X).\sigma_X=\sqrt{\sigma_X^2}=\sqrt{V(X)}.


Ex. XX follows Bernoulli(α)\text{Bernoulli}(\alpha). Calculate V(X)V(X). E(X)=αV(X)=E((x−2)2)=(0−α)2⋅(1−α)+(1−α)2⋅(α)=α2(1−α)+α(1−α)2=α(1−α)(α+(1−α))=α(1−α).\begin{aligned} E(X)&=\alpha\\ V(X)&=E((x-2)^2)\\ &=(0-\alpha)^2\cdot (1-\alpha)+(1-\alpha)^2\cdot (\alpha)\\ &=\alpha^2(1-\alpha)+\alpha(1-\alpha)^2\\ &=\alpha(1-\alpha)(\alpha+(1-\alpha))\\ &=\alpha(1-\alpha). \end{aligned}


Ex. XX follows Geo(p)\text{Geo}(p). Calculate V(X)V(X). E(X)=1pV(X)=∑k=1∞(k−1p)2⋅(1−p)k−1p=∑k=1∞(k2−2kp+1p2)⋅(1−p)k−1p=(∑k=1∞1p2(1−p)k−1p)−(∑k=1∞2kp(1−p)k−1p)+(∑k=1∞k2(1−p)k−1p)=1p2(∑k=1∞(1−p)k−1p)−2p(∑k=1∞k(1−p)k−1p)+p(∑k=1∞k2(1−p)k−1)=1p2(1)−2p(E(X))+p∑k=1∞k2(1−p)k−1=1p2−2p2+p∑k=1∞ddp(k(1−p)k)=−1p2+pddp(∑k=1∞k(1−p)k)=−1p2+pddp((1−p)∑k=1∞k(1−p)k−1)=−1p2+pddp((1−p)E(X))=−1p2+pddp(1p−1)=−(1p2+1p).\begin{aligned} E(X)&=\frac{1}{p}\\ V(X)&=\sum_{k=1}^\infin (k-\frac{1}{p})^2 \cdot (1-p)^{k-1}p\\ &=\sum_{k=1}^\infin (k^2-\frac{2k}{p}+\frac{1}{p^2}) \cdot (1-p)^{k-1}p\\ &=(\sum_{k=1}^\infin \frac{1}{p^2}(1-p)^{k-1}p) - (\sum_{k=1}^\infin \frac{2k}{p}(1-p)^{k-1}p) + (\sum_{k=1}^\infin k^2(1-p)^{k-1}p)\\ &=\frac{1}{p^2}(\sum_{k=1}^\infin (1-p)^{k-1}p) - \frac{2}{p}(\sum_{k=1}^\infin k(1-p)^{k-1}p) + p(\sum_{k=1}^\infin k^2(1-p)^{k-1})\\ &=\frac{1}{p^2}(1)-\frac{2}{p}(E(X))+p\sum_{k=1}^\infin k^2(1-p)^{k-1}\\ &=\frac{1}{p^2}-\frac{2}{p^2}+p\sum_{k=1}^\infin \frac{d}{dp} (k(1-p)^{k})\\ &=\frac{-1}{p^2}+p \frac{d}{dp} (\sum_{k=1}^\infin k(1-p)^{k})\\ &=\frac{-1}{p^2}+p \frac{d}{dp} ((1-p)\sum_{k=1}^\infin k(1-p)^{k-1})\\ &=\frac{-1}{p^2}+p \frac{d}{dp} ((1-p)E(X))\\ &=\frac{-1}{p^2}+p \frac{d}{dp} (\frac{1}{p}-1)\\ &=-(\frac{1}{p^2}+\frac{1}{p}). \end{aligned}


Proposition. V(X)=E(x2)−[E(X)]2.V(X)=E(x^2)-[E(X)]^2. V(X)=E[(x−μ)2]=E(x2−2μx+μ2)=E(x2)−E(2μx)+μ2=E(x2)−2μE(x)+μ2=E(x2)−2μ2+μ2=E(x2)−μ2=E(x2)−[E(X)]2.\begin{aligned} V(X)&=E[(x-\mu)^2]\\ &=E(x^2-2\mu x+\mu^2)\\ &=E(x^2)-E(2\mu x)+\mu^2\\ &=E(x^2)-2\mu E(x)+\mu^2\\ &=E(x^2)-2\mu^2 + \mu^2\\ &=E(x^2)-\mu^2\\ &=E(x^2)-[E(X)]^2. \end{aligned} We can use this fact to simplify the calculation of V(X)V(X).


Variance on Linear Functions

In general, given any function hh, what does V(h(X))V(h(X)) look like? V(h(X))=∑y∈D(h(y)−μh(X))2⋅p(y).V(h(X))=\sum_{y\in D}(h(y)-\mu_{h(X)})^2\cdot p(y). This doesn’t tell us much for the general case, since calculating μh(X)\mu_{h(X)} is not trivial.
However, when hh is a linear function (that is, h(x)=ax+bh(x)=ax+b), then h(y)−μax+b=ay+b−(aμX+b)=a(y−μX)h(y)-\mu_{ax+b}=ay+b-(a\mu_{X}+b)=a(y-\mu_X) and V(h(X))=V(ax+b)=∑y∈Da2(y−μX)2⋅p(y)=a2∑y∈D(y−μX)2⋅p(y)=a2V(X).\begin{aligned} V(h(X))&=V(ax+b)\\ &=\sum_{y\in D} a^2(y-\mu_X)^2 \cdot p(y)\\ &=a^2\sum_{y\in D} (y-\mu_X)^2 \cdot p(y)\\ &=a^2V(X). \end{aligned}

Thus, V(ax+b)=σax+b2=a2σX2.V(ax+b)=\sigma_{ax+b}^2=a^2\sigma_{X}^2.