A police officer is measuring the speed of cars on a high way. The actual speed X of a car is $\mu(80,120)$. Due to the inaccuracy of the speed gun, the speed Y measured,given X=x, is $N (x,\frac{1}{100}x)$. Find the mean/ variance of Y.

The answer is $E(Y)=E[E(Y|X)]=E[X]=100$ by the law of total expectation.

By the law of total Var: \begin{align} Var(Y)= & E(Var(Y|X))+Var(E(Y|X)) \\ =&E(\frac{1}{100}x)+Var(x) \\ =& 1+\frac{400}{3} \end{align}

Question: for the algorithm in computing these, I am not familiar, can someone explain the steps missed in the middle for both expectation and variance. Appreciate it.


1 Answer 1


I presume the symbol $\mu$ was meant to indicate that $X\sim\mathcal{U}(80..120)$, that is uniformly distributed, so ...$$\mathsf E(X)=100~\\\mathsf{Var}(X)=400/3$$

You were also told $Y\mid X\sim\mathcal N(X,X/100)$, that is the conditional distribution of $Y$ given $X$ is Normal, with ... $$\mathsf E(Y\mid X)=X~~~~\\\mathsf {Var}(Y\mid X)= X/100$$

Everything else is just the Laws of Total Expectation and of Total Variance, and substitution.

$$\begin{align} \mathsf E(Y)&=\mathsf E(\mathsf E(Y\mid X))&~:~&\text{Law of Total Expectation} \\[1ex]&=\mathsf E(X)&~:~&\mathsf E(Y\mid X)=X \\[1ex]&=100&~:~&\mathsf E(X)=100 \\[4ex] \mathsf {Var}(Y)&=\mathsf E(\mathsf {Var}(Y\mid X))+\mathsf {Var}(\mathsf E(Y\mid X))&~:~&\text{Law of Total Variance}\\[1ex]&=\mathsf E(X/100)+\mathsf {Var}(\mathsf E(Y\mid X))&~:~&\mathsf{Var}(Y\mid X)=X/100\\[1ex]&=\mathsf E(X/100)+\mathsf {Var}(X)&~:~&\mathsf E(Y\mid X)=X\\[1ex]&=100/100+\mathsf{Var}(X)&~:~&\mathsf E(X)=100\\[1ex]&=1+400/3&~:~&\mathsf {Var}(X)=400/3\\[1ex]&=134.\dot 3\end{align}$$


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