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STT 861 Theory of Prob and STT I Lecture Note - 10


Solving a problem in midterm exam; conditional distribution, definition, example and its expectation and variance.

Portal to all the other notes

Lecture 10 - Nov 08 2017

Exam question

Let $X\sim U(50,90)$, $Z\sim U(70,90)$, $Law(Z) = Law(X|X\geq 70)$.

Q: Let $X\sim U(a,b)$, $c\in (a,b)$, find law ($X|X>c$).

A: let $F$ be the CDF of $X|X>c$. Let $x\in [c,b]$.

\[F(x)=P(X\leq x|x>c)=\frac{P(X\leq x, X|c)}{P(X>c)}=\frac{\frac{x-c}{b-a}}{\frac{b-c}{b-a}}=\frac{x-c}{b-c}\]

It is uniform in $(c, b)$.

Conditional distribution - Chapter 5

Recall joint density of a vector $(X,Y)$ is $f_{X,Y}(x,y)$, $x$ and $y$ are real numbers.

Definition (Theorem): The conditional law of $X$ given $Y$ is

\[f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}\]

Consequently, the CDF of $X$ given $Y=y$ is

\[F_{X|Y}=P(X\leq x|Y= y)=\int_{-\infty}^{x}f_{X|Y}(u,v)du\]

Discrete case using the same idea.

For $Y$, no change in notation.

For $X$, use $\sum$ instead of integrals.

Example 1

Let $N\sim Poi(\lambda)$ and $X\sim Bin(N,\theta)$, here we see law $(X|N=k)=Bin(k,\theta)$.

This type of example is called a mixture. A mixture is when the parameter of one random variable is determined by the value of another random variable.

In our example, we would like to find law $(X)$ unconditional on $N$. We want to find $P(X=i)$ for any $k=0,1,2,…$.

\[\begin{align*} P(X=i) &= \sum_{k=i}^{\infty}P(X=i ~and~N=k) \\ &= \sum_{k=i}^{\infty} P(X=i|N=k)P(N=k) \\ &= \sum_{k=i}^{\infty} C_k^i \theta^i(1-\theta)^{k-i} e^{-\lambda}\frac{\lambda^k}{k!} \\ &= \sum_{k=i}^{\infty}\frac{k!}{i!(k-i)!}\theta^i(1-\theta)^{k-i}e^{-\lambda}\frac{\lambda^k}{k!} \\ &= \frac{e^{-\lambda}\theta^i}{i!}\sum_{k=i}^{\infty}\frac{(1-\theta)^{k-i}}{(k-i)!}\lambda^k \\ &= \frac{e^{-\lambda}\theta^i}{i!}\sum_{k'=0}^{\infty} \frac{(i-\theta)^{k'}}{k'!}\lambda^{k'+i} \\ &= \frac{e^{-\lambda}(\theta\lambda)^i}{i!}\sum_{k'=0}^{\infty} \frac{(i-\theta)^{k'}}{k'!}\lambda^{k'} \\ &= \frac{e^{-\lambda}(\theta\lambda)^i}{i!}e^{\lambda(1-\theta)} \\ &= e^{-\lambda\theta}\frac{(\lambda\theta)^i}{i!} \end{align*}\]

It is Poisson distribution with parameter $\lambda’=\lambda\theta$.

Home exercise

Read and understand the section of the book in Chapter 5 on Markov Chains.

Conditional expectations - Chapter 5.2


For the discrete case,


Super important theorem (definition)

let $X$ and $Y$ be given. Let

\[g(x)=E(Y|X=x)\] \[v(x)=Var(Y|X=x)=E(Y^2|X=x)-(E(Y|X=x))^2 = E((Y-g(x))^2|X=x)\]

Theorem: $E(XY) = E(Xg(x))$

(the idea of “tower property”: $E(E(Z|X))=E(Z)$. “Given $X$” means $X$ is not random).

Proof: use the property to prove the theorem

\[\begin{align*} E(XY) &= E(E(XY|X)) \\ &= E(XE(Y|X)) \\ &= E(Xg(X)) \end{align*}\]


Now recall $v(x)$. It turns out

\[Var(Y)=E(v(X)) + Var(g(X)) = E(v(x))+var(E(Y|X))\]

Explanation: The unconditional variance of $Y$ is the expected conditional variance plus the variance of the conditional expectation.

Example 2

Let $X|N \sim Bin (N,\theta)$, $N\sim Poi(\lambda)$.

(We found that $X\sim Poi(\lambda)$)

From the theorem, ignoring the fact above, we apply $E(X)=E(E(X|N))$

If we give some information and then take the information away, it is equivalent to not giving information at the first place.

\[E(X)=E(E(X\|N))=\theta E(N)=\lambda\theta\] \[E(Poi(\lambda\theta))=\lambda\theta\]

We verify the theorem above.

For variance

\[Var(X)=E(v(N))+Var(E(X|N)) = N\theta(1-\theta) + N\theta=\theta(1-\theta)\lambda+\theta^2\lambda = \lambda\theta\]

This is also correct.

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