Tower property
Theorem taught in Joint distributions, part of Probability
Tower property
| Type | Theorem |
| Taught in | Joint distributions |
| Part | Probability |
| Worked examples | 2 |
If the expectations exist, then \mathbb{E}\bigl(\mathbb{E}(Y\mid X)\bigr)=\mathbb{E} Y. More generally, \mathbb{E}\bigl(\mathbb{E}(Y\mid X,Z)\mid X\bigr)=\mathbb{E}(Y\mid X).
Examples
A uniform point on a triangle
Worked example from Joint distributions
A point (X,Y) is drawn uniformly from the 10 lattice points with 1\le x\le y\le4, so each has probability 1/10.
Question. Find the marginals, the conditional distribution of Y given X=x, and \mathbb{E}(Y\mid X). Check the tower property, and find the correlation.
Solution. For a given x there are 5-x admissible values of y, and for a given y there are y values of x, so
f_X(x)=\frac{5-x}{10},\qquad f_Y(y)=\frac{y}{10},\qquad x,y\in\{1,2,3,4\}.
X and Y are dependent: given X=2, Y can only be 2, 3 or 4. The conditional mass function is
f_{Y\mid X}(y\mid x)=\frac{1/10}{(5-x)/10}=\frac1{5-x},\qquad y=x,\ldots,4,
uniform on the admissible values. So \mathbb{E}(Y\mid X=x)=(x+4)/2. Averaging over X,
\mathbb{E}\bigl(\mathbb{E}(Y\mid X)\bigr)=\tfrac52\cdot\tfrac4{10}+3\cdot\tfrac3{10}+\tfrac72\cdot\tfrac2{10}+4\cdot\tfrac1{10}=3,
which agrees with \mathbb{E} Y=(1+4+9+16)/10=3. For the correlation, \mathbb{E} X=2, \mathbb{E}(XY)=13/2 and both variances are 1, so \operatorname{Cov}(X,Y)=13/2-6=1/2 and \rho(X,Y)=1/2. A larger X forces a larger Y.
A day of takings
Worked example from Joint distributions
A shop has N\sim\operatorname{Poisson}(20) customers a day. Customers spend independently of one another and of N, each with mean 30 dollars and standard deviation 10 dollars.
Question. Find the mean and standard deviation of the day’s takings.
Solution. The takings are a random sum Y=X_1+\cdots+X_N. By the tower property, \mathbb{E} Y=\mu\,\mathbb{E} N=30\times20=600 dollars. For the variance, \operatorname{Var}(N)=20 for a Poisson count, so
\operatorname{Var}(Y)=\mathbb{E}(N)\operatorname{Var}(X_1)+\operatorname{Var}(N)(\mathbb{E} X_1)^2=20\times100+20\times900=20000,
and the standard deviation is 141 dollars. Most of the variability comes from the number of customers, not from what each one spends: if N were fixed at 20, the standard deviation would be only \sqrt{2000}=45 dollars.
Used in
- Taught in Joint distributions, where the derivation and examples are.
- All 8 questions on Joint distributions
See also
Expectation · Joint probability mass function · Independent random variables · Joint probability density · Covariance · Expectation vector and covariance matrix · Conditional distribution