The price of a painting is @@@\$100@@@. In two years it will be worth @@@5000@@@ with probability @@@2\%@@@ and nothing with probability @@@98\%@@@. What is the expected value of the painting in two years?

Our RV @@@X@@@, representing the price of the painting in two years, is equal to @@@5000@@@ with probability @@@0.02@@@ and equal to @@@0@@@ with probability @@@.98@@@. Therefore the expected value is equal to

$$ E[X] = 5000*0.02 + 0*0.98 = 100.$$

Just the same as today. Maybe but another painting as an investment or just buy this because we like it.

In a jar I have @@@3@@@ white balls and @@@4@@@ black balls. I select two balls. What is the expected number of white balls?

Let @@@X@@@ be the number of white balls. Then @@@X@@@ is hypergeometric. Thus, the PMF @@@p_X@@@ is given by

$$\begin{equation*} \begin{array}{|r|c|c|c|} \hline\hline x = & 0 & 1 & 2\\ \hline p_X(x) = & \frac{ \binom{4}{2}}{\binom{7}{2}} & \frac{\binom{4}{1}\binom{3}{1}}{\binom{7}{2}} & \frac{\binom{3}{2}}{\binom{7}{2}} \\ \hline x* p_X(x) = & 0*\frac{6}{21} & 1*\frac{12}{21} & 2*\frac{3}{21} \\ \hline\hline \end{array} \end{equation*}$$

The expectation is the sum of the numbers in the last row, and is therefore equal to @@@\frac{18}{21}=\frac{6}{7}@@@. This is a little short of @@@1@@@, because there are less white balls than black balls. But wait! Let @@@X_1@@@ be thed indicator of first ball is white, and let @@@X_2@@@ be the indicator that the second ball is white. Then we know that both are indicators of events, each having probability @@@3/7@@@ (using the formula we derived when discussing independence), therefore @@@X_1,X_2\sim \mbox{Bern}(3/7)@@@ and @@@E[X_1]=E[X_2]=3/7@@@. Also, @@@X=X_1+X_2@@@, and our calculation above shows that @@@E[X]=E[X_1]+E[X_2]@@@. Is this just a coincidence? No. This is an illustration of one of the most important properties of the expectation, see Theorem Theorem 3.

1. ’’‘Continuous Uniform’’’: @@@X\sim \mbox{U}[a,b]@@@. Then @@@E[X] = \frac{a+b}{2}@@@. Indeed,

$$E[X] = \frac{1}{b-a} \int_a^b u du = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}{2}.$$

1. ’’‘Exponential’’’: @@@X\sim \mbox{Exp}(\lambda)@@@. Then @@@E[X] = \frac{1}{\lambda}@@@. Indeed,

$$ E[X] = \int_0^\infty u \lambda e^{-\lambda u} du=(*)$$

We can calculate this through integration by parts:

$$ (*) = u (-e^{-\lambda u})|_0^\infty +\int_0^\infty e^{-\lambda u} du = \frac{1}{\lambda}.$$

Alternatively, we can use the following differentiation under integral sign trick

$$ (*) = \int_0^\infty \lambda \frac{\partial }{\partial \lambda} (-e^{-\lambda u}) d u = -\lambda\frac{\partial }{\partial \lambda} \int_0^\infty e^{-\lambda u}du = -\lambda \frac{\partial }{\partial \lambda} \frac{1}{\lambda}= \frac{1}{\lambda}.$$

Or, we can simply use Definition Definition 2:

$$ E[X] = \int_0^\infty P(X>t) dt - 0 = \int_0^\infty 1-F_X(t) dt = \int_0^\infty e^{-\lambda t} dt = \frac{1}{\lambda}.$$

It’s all a matter of taste, but you need to suppose now your integrals.

I open the store at @@@8\text{AM}@@@. The first customer arrives to the store @@@X@@@ hours after its opening where @@@\mbox{U}[0,2]@@@, and stays there for @@@Y@@@ hours, where @@@Y\sim \mbox{U}[0,2]@@@ What is the expected time the first customer leaves the store?

We need to find @@@E[ 8+ X+Y]@@@. The Proposition gives

$$ E[8+X+Y] = E[8]+E[X]+E[Y] = 8+ 1+1= 10.$$

The expected time the first customer departs is @@@10:00\text{AM}@@@. It is utterly important to note that in this problem it is utterly wrong to try to calculate the CDF or density of @@@X+Y@@@ because the data do not determine it uniquely. Why? For example, @@@Y@@@ may be equal to @@@X@@@ (a customer who chooses to stay an amount equal to time of arrival after store opening for whatever reason), or @@@Y@@@ can be @@@2-X\sim \mbox{U}[0,2]@@@ (the customer decides to leave at time @@@10:00@@@). In the first case @@@X+Y=2X\sim\mbox{U}[0,4]@@@. In the latter case, @@@X+Y = 2@@@, a constant RV, hence discrete. The point is that the linearity of expectation allows to completely avoid this discussion: it’s irrelevant.

Let’s revisit Example Example 2. Let @@@X_1@@@ and @@@X_2@@@ be the indicators of the first ball being white and second being white, respectively. Note, that the events are not independent, but have the same probability, equal to @@@\frac{\binom{3}{1}}{\binom{7}{1}} = \frac{3}{7}@@@, and therefore @@@E[X_1]=E[X_2]= \frac{3}{7}@@@. Now @@@X= X_1+X_2@@@. Therefore @@@E[X]=E[X_1]+E[X_2] = \frac{6}{7}@@@. Let’s suppose we select four balls. What is then the expectation of the number of white balls? By linearity of expectation it would be @@@\frac{4*3}{7}=\frac{12}{7}@@@. Finally, sanity check: what if we select all balls? Then clearly we have all @@@3@@@ white balls, so the answer is @@@3@@@, and indeed, this is equal to @@@7*\frac{3}{7}@@@.

In a barn, @@@100@@@ chicks sit peacefully in a circle. Suddenly, each chick randomly pecks the chick immediately to its left or right. What is the expected number of unpecked chicks? source

Before answering, let’s generalize to make it more interesting. Suppose each chick chooses a side with probability @@@p@@@ to be clockwise and @@@1-p@@@ for counterclockwise, and then pecks the chick to that side with probability @@@q@@@. In the original problem @@@p=1/2@@@ and @@@q=1@@@.

Label the chicks as @@@1,\dots,100@@@, and let @@@X_i@@@ be the indicator of chick @@@i@@@ not being pecked. Then @@@X_i@@@ is equal to @@@1@@@ if neither the clockwise neighbor nor the counterclockwise neighbor pecked it.

• The event that the counterclockwise neighbor does not peck it is @@@p(1-q)+(1-p)@@@ (choosing that side, but deciding not to peck, or not choosing that side).
• Similarly, the probability that the clockwise neighbor does not peck is @@@(1-p)(1-q)+p@@@.

Since each of the events is determined by a different chick, they are independent. And so their intersection, the event that our own chick is not pecked, has probability @@@(p(1-q)+(1-p))((1-p)(1-q)+p)=(1-pq)(1-(1-p)q)@@@. Since @@@X_i@@@ is the indicator of this event, this number is also the expectation of @@@X_i@@@. Since the number of chicks not pecked is @@@X_1+\dots+X_100@@@, it follows from the additivity of expectation that the expected number of those left not pecked is @@@100 (1-pq)(1-(1-p)q)@@@. Note that this can be as small as @@@0@@@ (@@@p=1,q=1@@@) or as high as @@@100@@@ (@@@q=0@@@). In the case @@@p=1/2@@@ and @@@q=1@@@ we get @@@25@@@.

I’m shooting my basketball @@@n\ge 6@@@ times. The probability I make a shot is @@@p@@@, and all shots are independent. What is the expected number of streaks of @@@6@@@ consecutive successful shots? source

Let @@@X_j,~j=1,\dots,n-5@@@ be the indicator of the event that a streak of @@@6@@@ successful shots begins at the @@@j@@@-th shot. Clearly @@@E[X_j] = P(X_j=1) =p^6@@@ for all @@@j@@@. The number of streaks is equal to @@@X_1+ \dots + X_{n-5}@@@. Therefore the expected number of streaks is @@@(n-5)p^6@@@.

• ’’‘Bernoulli’’’: Suppose @@@X\sim \mbox{Bern}(p)@@@. Then @@@E[X] = 1 *p + 0*(1-p)=p@@@.
• ’’‘Binomial’’’: Suppose @@@X \sim \mbox{Bin}(n,p)@@@. Then @@@X= \sum_{j=1}^n X_j@@@, where @@@X_1,\dots,X_n@@@ are each @@@\mbox{Bern}(p)@@@. By linearity of expectation

$$E[X]= E[X_1]+\dots + E[X_n]=np.$$

• ’’‘Geometric’’‘:Suppose @@@X\sim \mbox{Geom}(p)@@@. Then

$$\begin{align*} E[X] &= \sum_{k=1}^\infty k x^{k-1} p,~\mbox{ where } x=1-p \\ & = p\sum_{k=1}^\infty \frac{d}{dx} x^k \\ & = p \frac{d}{dx} \frac{x}{1-x} \\ & p \frac{ d}{dx} (-1 + \frac{1}{1-x})\\ & p \frac{1}{(1-x)^2} = \frac{1}{p}. \end{align*}$$

Want another idea? Let’s use Definition Definition 2:

$$\begin{align*} E[X]& = \int_0^\infty P(X>t) dt\\ & = \sum_{j=0}^\infty \int_j^{j+1} P(X>t) dt = \sum_{j=0}^\infty P(X>j)\\ & = \sum_{j=0}^\infty (1-p)^j= \frac{1}{1-(1-p)}=\frac 1p. \end{align*}$$

Here’s one more… @@@X@@@ represents the time of the first success:

1. With probability @@@p@@@ we’re done at the first trial, and with probability @@@1-p@@@ the first trial fails so we need to continue.
2. @@@\Rightarrow@@@ we count at least one trial with probability @@@1@@@, and;
3. with probability @@@1-p@@@ add the number of trials until the first success, but starting from the second trial. The expected number of trials (counting from the second trial) until the first success is, again, @@@E[X]@@@.

Putting both together, we have

$$E[X] = 1+ (1-p) E[X],$$

and the only finite solution is @@@E[X]=\frac{1}{p}@@@.

Okay, but that may not seem rigorous enough, right? We will be able to justify this line of thinking in a nice and clean way when we deal with conditional expectation. In the meantime, let’s try to do this by manipulating power series. Try to identify the ideas described above as steps in the calculation:

$$\begin{align*} E[X] & = \sum_{k=1}^\infty k (1-p)^{k-1}p \\ & = 1*p + \sum_{k=2}^\infty k (1-p)^{k-1}p \\ & = 1*p + (1-p) \sum_{k=2}^\infty k (1-p)^{k-2}p\\ & = 1*p +(1-p) \sum_{j=1}^\infty (j+1) (1-p)^{j-1}p\\ & = 1*p + (1-p) \left ( \underset{=E[X]}{\underbrace{\sum_{j=1}^\infty j (1-p)^{j-1}p}} + \underset{=1}{\underbrace{\sum_{j=1}^\infty (1-p)^{j-1}p}}\right)\\ & = 1+ (1-p) E[X] \end{align*}$$

• ’’‘Hypergeometric’’’: A brute force calculation is not hard, but let’s rely on additivity of expectation. Recall: we’re selecting (without replacement) @@@n@@@ balls from a jar that contains @@@N@@@ balls, @@@K@@@ of which are white and @@@b=N-K@@@ are black, and @@@X@@@ the number of white balls selected. If @@@X_1,\dots,X_n@@@ are the indicators that the first, second,… @@@n@@@-th ball is white, then these @@@n@@@ RVs are @@@\mbox{Bern}(K/N)@@@ (we showed this when studying independence), and thus each has expectation @@@K/N@@@. It follows from the additivity of the expectation that

$$E[X]= \sum_{j=1}^n E[X_j] = n \frac{K}{N}.$$

• ’’’ Negative Binomial’’’: Negative binomial with parameters @@@r@@@ and @@@p@@@, @@@X\sim \mbox{NB}(r,p)@@@. We take the liberty not to do the direct calculation. Recall that @@@X@@@ counts the number of failures until the @@@r@@@-th success in a sequence of independent trials where the probability of success in each is @@@1-p@@@. If @@@X_r@@@ is the number of trials until the @@@r@@@-th success, then clearly @@@X=X_r-r@@@, while @@@X_r@@@ is a sum of @@@r@@@ @@@\mbox{Geom}(1-p)@@@ RVs (number of trials between any two successive successes is @@@\mbox{Geom}(1-p)@@@. Therefore,

$$ E[X] = E[X_r - r]= \frac{r}{1-p}-r = \frac{r p}{1-p}.$$

• ’’‘Poisson’’’: Suppose @@@X\sim \mbox{Pois}(\lambda)@@@. Then

$$\begin{align*}E[X] &= \sum_{k=0}^\infty ke^{-\lambda}\frac{ \lambda^k}{k!} \\ & = \lambda \sum_{k=0}^\infty e^{-\lambda} \frac{ \lambda^{k-1}}{(k-1)!}=\lambda \times 1.\end{align*}$$

• ’’‘Discrete Uniform’’’: Suppose that @@@X@@@ is uniform on @@@\{1,\dots,n\}@@@. Then

$$E[X] = \frac{1}{n} \underset{\star}{\underbrace{\sum_{j=1}^n j}}.$$

Now @@@\sum_{j=1}^n j = \sum_{j=1}^n (n+1-j)@@@. Therefore

$$\star= \frac 12 \left (\sum_{j=1}^n j + \sum_{j=1}^n (n+1-j)\right)= \frac 12 n \times (n+1),$$

and so po

$$E[X] = \frac{n(n+1)}{2 n} = \frac{n+1}{2}.$$