- The CDF is defined by $F_X(x)=P(X \leq x)$. We have \begin \nonumber F_X(x) = \left\< \begin0 & \quad \text x
Problem
- Find $EX$.
- Find Var$(X)$.
- If $Y=(X-2)^2$, find $EY$.
| $EX$ | $= \sum_ x_kP_X(x_k)$ |
| $= 0 (0.1)+ 1(0.4)+2(0.3)+3(0.2)$ | |
| $=1.6$ |
Let $X$ be a discrete random variable with PMF \begin \nonumber P_X(k) = \left\< \begin 0.2 & \quad \text k=0\\ 0.2 & \quad \text k=1\\ 0.3 & \quad \text k=2\\ 0.3 & \quad \text k=3\\ 0 & \quad \text \end \right. \end Define $Y=X(X-1)(X-2)$. Find the PMF of $Y$.
| $P_Y(0)$ | $=P(Y=0)=P\big( (X=0) \textrm < or >(X=1) \textrm < or >(X=2)\big)$ |
| $=P_X(0)+P_X(1)+P_X(2)$ | |
| $=0.7$; | |
| $P_Y(6)$ | $= P(X=3)=0.3$ |
Let $X \sim Geometric(p)$. Find $E\left[\frac\right]$.
- Solution
- The PMF of $X$ is given by \begin \nonumber P_X(k) = \left\< \beginpq^& \quad \text k=1,2,3. \\ 0 & \quad \text \end \right. \end where $q=1-p$. Thus,
$E\left[\frac\right]$ $=\sum_^ <\infty>\frac P_X(k)$ $=\sum_^ <\infty>\frac q^p$ $=\frac \sum_^ <\infty>\left(\frac\right)^$
$=\frac \frac<1-\frac>$
$=\frac $.
If $X \sim Hypergeometric(b,r,k)$, find $EX$.
- Solution
- The PMF of $X$ is given by \begin \nonumber P_X(x) = \left\< \begin\frac >>& \quad \text x \in R_X\\ 0 & \quad \text \end \right. \end where $R_X=\<\max(0,k-r), \max(0,k-r)+1, \max(0,k-r)+2. \min(k,b)\>$. Finding $EX$ directly seems to be very complicated. So let's try to see if we can find an easier way to find $EX$. In particular, a powerful tool that we have is linearity of expectation. Can we write $X$ as the sum of simpler random variables $X_i$? To do so, let's remember the random experiment behind the hypergeometric distribution. You have a bag that contains $b$ blue marbles and $r$ red marbles. You choose $k \leq b+r$ marbles at random (without replacement) and let $X$ be the number of blue marbles in your sample. In particular, let's define the indicator random variables $X_i$ as follows: \begin \nonumber X_i = \left\< \begin1 & \quad \text \\ 0 & \quad \text \end \right. \end Then, we can write $$X=X_1+X_2+\cdots+X_k.$$ Thus, $$EX=EX_1+EX_2+\cdots+EX_k.$$ To find $P(X_i=1)$, we note that for any particular $X_i$ all marbles are equally likely to be chosen. This is because of symmetry: no marble is more likely to be chosen than the $i$th marble as any other marbles. Therefore, $$P(X_i=1)=\frac\textrm< for all >i \in \.$$ We conclude
$EX_i$ $=0 \cdot p(X_i=0)+ 1 \cdot P(X_i=1)$ $=\frac$. In Example 3.14 we showed that if $X \sim Binomial(n,p)$, then $EX=np$. We found this by writing $X$ as the sum of $n$ $Bernoulli(p)$ random variables. Now, find $EX$ directly using $EX=\sum_ x_k P_X(x_k)$. Hint: Use $k =n $.
- Solution
- First note that we can prove $k =n $ by the following combinatorial interpretation: Suppose that from a group of $n$ students we would like to choose a committee of $k$ students, one of whom is chosen to be the committee chair. We can do this
- by choosing $k$ people first (in $$ ways), and then choosing one of them to be the chair ($k$ ways), or
- by choosing the chair first ($n$ possibilities and then choosing $k-1$ students from the remaining $n-1$ students (in $$ ways)).
$EX$ $=\sum_^ k p^k q^$ $=\sum_^ k p^k q^$ $=\sum_^ n p^k q^$ $=np\sum_^ p^ q^$ $=np\sum_^ p^l q^$ $=np$.
Problem
Let $X$ be a discrete random variable with $R_X \subset \$. Prove $$EX=\sum_^ <\infty>P(X>k).$$$P(X > 0)$ $=P_X(1)+P_X(2)+P_X(3)+P_X(4)+\cdots$, $P(X > 1)$ $=P_X(2)+P_X(3)+P_X(4)+\cdots$, $P(X > 2)$ $=P_X(3)+P_X(4)+P_X(5)+\cdots$. $\sum_^ <\infty>P(X>k)$ $= P(X>0)+P(X>1)+P(X>2)+. $ $=P_X(1)+2P_X(2)+3P_X(3)+4P_X(4)+. $ $=EX$. If $X \sim Poisson(\lambda)$, find Var$(X)$.
- Solution
- We already know $EX=\lambda$, thus Var$(X)=EX^2-\lambda^2$. You can find $EX^2$ directly using LOTUS; however, it is a little easier to find $E[X(X-1)]$ first. In particular, using LOTUS we have
$E[X(X-1)]$ $=\sum_^ <\infty>k(k-1)P_X(k)$ $=\sum_^ <\infty>k(k-1) e^ <-\lambda>\frac<\lambda^k>$ $=e^ <-\lambda>\sum_^ <\infty>\frac<\lambda^k>$ $=e^ <-\lambda>\lambda^2 \sum_^ <\infty>\frac<\lambda^ >$ $=e^ <-\lambda>\lambda^2 e^<\lambda>$ $=\lambda^2$.
So, we have $\lambda^2=E[X(X-1)]=EX^2-EX=EX^2-\lambda$. Thus, $EX^2=\lambda^2+\lambda$ and we conclude$\textrm(X)$ $=EX^2-(EX)^2$ $=\lambda^2+\lambda-\lambda^2$ $=\lambda$. Let $X$ and $Y$ be two independent random variables. Suppose that we know Var$(2X-Y)=6$ and Var$(X+2Y)=9$. Find Var$(X)$ and Var$(Y)$.
- Solution
- Let's first make sure we understand what Var$(2X-Y)$ and Var$(X+2Y)$ mean. They are Var$(Z)$ and Var$(W)$, where the random variables $Z$ and $W$ are defined as $Z=2X-Y$ and $W=X+2Y$. Since $X$ and $Y$ are independent random variables, then $2X$ and $-Y$ are independent random variables. Also, $X$ and $2Y$ are independent random variables. Thus, by using Equation 3.7, we can write $$\textrm (2X-Y)=\textrm(2X)+\textrm=4\textrm(X)+\textrm=6,$$ $$\textrm (X+2Y)=\textrm(X)+\textrm=\textrm(X)+4\textrm=9.$$ By solving for Var$(X)$ and Var$(Y)$, we obtain Var$(X)=1$ and Var$(Y)=2$.
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Practical uncertainty: Useful Ideas in Decision-Making, Risk, Randomness, & AI
- First note that we can prove $k =n $ by the following combinatorial interpretation: Suppose that from a group of $n$ students we would like to choose a committee of $k$ students, one of whom is chosen to be the committee chair. We can do this