Discrete random variables

Worked example 1 — From outcomes to numerical values

Two fair coins are tossed independently. Let \(X\) be the number of heads in the two tosses. Find the probability distribution of \(X\).

Solution

Write H for heads and T for tails. The equally likely outcomes are TT, TH, HT and HH, each with probability \(\frac14\). Their values of \(X\) are 0, 1, 1 and 2.

Thus \(\Prob(X=1)=\frac12\): two of the four equally likely outcomes produce one head. The three possible values are not equally likely.

Worked example 2 — Find a normalising constant

Let \(X\) be the number of tokens in a randomly selected envelope, with \[\begin{gathered}\Prob(X=x)=k(x+1),\\x\in\{1,2,3\}.\end{gathered}\] and probability zero for all other values. Find \(k\) and check that the distribution is valid.

Solution

The normalising constant \(k\) makes the probabilities add to 1: \[\begin{aligned}k(1+1)&+k(2+1)\\&+k(3+1)=1,\\9k&=1,\\k&=\frac19.\end{aligned}\]

Each entry lies between 0 and 1; their sum is \(\frac29+\frac39+\frac49=1\). Both conditions hold.

The stem at each \(x\) has height \(\Prob(X=x)\). Values outside the table have probability zero.

Worked example 3 — Inequalities and endpoints

Let \(X\) be the change in a player’s score, in points, after one round:

Find:

1. \(\Prob(X>1)\);

2. \(\Prob(X\geq1)\);

3. \(\Prob(X\leq0)\), using the complement.

Solution

1. Only the value 3 is greater than 1: \[\begin{aligned}\Prob(X>1)&=\Prob(X=3)\\&=\boxed{\frac3{10}}.\end{aligned}\]

2. Both 1 and 3 are at least 1: \[\Prob(X\geq1)=\frac4{10}+\frac3{10}=\boxed{\frac7{10}}.\]

3. The complement of \(X\leq0\) is \(X>0\). The positive values in this table are 1 and 3: \[\begin{aligned}\Prob(X\leq0)&=1-\Prob(X>0)\\&=1-\left(\frac4{10}+\frac3{10}\right)\\&=\boxed{\frac3{10}}.\end{aligned}\] Here \(\Prob(X>0)=\Prob(X\geq1)\) because there are no possible values between 0 and 1.

Worked example 4 — A probability after restricting the outcomes

Use the score-change distribution from example 3:

Given that the player’s score did not decrease, find the probability that it increased by at least 1 point. In symbols, find \(\Prob(X\geq1\mid X\geq0)\).

Solution

Identify the condition. We are told that \(X\geq0\), so the remaining values are 0, 1 and 3. Their total probability is \[\begin{aligned}\Prob(X\geq0)&=\frac2{10}\\&\quad{}+\frac4{10}\\&\quad{}+\frac3{10}\\&=\frac9{10}.\end{aligned}\] Identify the favourable values within that condition. The values 1 and 3 satisfy both \(X\geq1\) and \(X\geq0\): \[\begin{gathered}\Prob\bigl((X\geq1)\cap(X\geq0)\bigr)\\=\frac4{10}+\frac3{10}\\=\frac7{10}.\end{gathered}\] Divide by the probability of the condition: \[\Prob(X\geq1\mid X\geq0)=\frac{7/10}{9/10}=\boxed{\frac79}.\] The probability is \(\frac79\) among trials in which the score did not decrease.

Worked example 5 — A prize-envelope model

Let \(X\) be the token count in a randomly selected envelope from a new supply:

Find and interpret the expected number of tokens.

Solution

\[\begin{aligned}\E(X)&=0\left(\frac15\right)\\&\quad{}+2\left(\frac12\right)\\&\quad{}+5\left(\frac3{10}\right)\\&=\boxed{\frac52\,\mathrm{tokens}}.\end{aligned}\] Over 1000 independent selections, counts of about 200, 500 and 300 would give an average of \[\begin{gathered}\frac{200(0)+500(2)+300(5)}{1000}\\=2.5\,\mathrm{tokens}.\end{gathered}\] The actual counts fluctuate. No individual envelope contains 2.5 tokens.

Worked example 6 — A mean can determine missing probabilities

The possible scores \(X\) in a game are 0, 1 and 3 points, with probabilities \(a\), \(b\) and \(\frac14\). Given that \(\E(X)=\frac54\), find \(a\) and \(b\).

Solution

Use two facts: total probability is 1, and the mean is \(\frac54\): \[\begin{aligned} a+b+\frac14&=1,\\ 0a+1b+3\left(\frac14\right)&=\frac54. \end{aligned}\] The mean equation gives \(b=\frac54-\frac34=\frac12\). Then \(a=1-\frac14-\frac12=\frac14\).

Worked example 7 — Spread of the token count

Use the envelope distribution from example 5: \(X=0,2,5\) tokens with probabilities \(\frac15,\frac12,\frac3{10}\). We already know \(\E(X)=\frac52\) tokens. Find:

1. \(\E(X^2)\);

2. the variance of \(X\);

3. the standard deviation of \(X\).

Solution

1. Square each value and multiply by its probability: \[\begin{aligned}\E(X^2)&=0^2\left(\frac15\right)\\&\quad{}+2^2\left(\frac12\right)\\&\quad{}+5^2\left(\frac3{10}\right)\\&=\frac{19}{2}\,\mathrm{tokens}^2.\end{aligned}\]

2. Subtract the square of the mean: \[\begin{aligned}\Var(X)&=\frac{19}{2}-\left(\frac52\right)^2\\&=\boxed{\frac{13}{4}\,\mathrm{tokens}^2}.\end{aligned}\]

3. Take the square root of the variance: \[\begin{aligned}\sigma&=\sqrt{\frac{13}{4}}\\&=\boxed{\frac{\sqrt{13}}2\,\mathrm{tokens}}\\&\approx1.80\,\mathrm{tokens}.\end{aligned}\]

Worked example 8 — Same mean, different spread

Two machines label packs containing two items. Let \(A\) and \(B\) be the number of unlabelled items in a pack from machines A and B, respectively.

Compare the means, variances and standard deviations of \(A\) and \(B\).

Solution

\[\begin{aligned}\E(A)&=0(0.1)\\&\quad{}+1(0.8)\\&\quad{}+2(0.1)\\&=1,\\\E(B)&=0(0.4)\\&\quad{}+1(0.2)\\&\quad{}+2(0.4)\\&=1.\end{aligned}\] Using \(\Var(X)=\E(X^2)-[\E(X)]^2\), \[\begin{aligned}\Var(A)&=0^2(0.1)\\&\quad{}+1^2(0.8)\\&\quad{}+2^2(0.1)\\&\quad{}-1^2\\&=0.2,\\\Var(B)&=0^2(0.4)\\&\quad{}+1^2(0.2)\\&\quad{}+2^2(0.4)\\&\quad{}-1^2\\&=0.8.\end{aligned}\] \[\begin{aligned} \sigma_A&=\sqrt{0.2}\approx0.447\,\mathrm{items},\\ \sigma_B&=\sqrt{0.8}\approx0.894\,\mathrm{items}. \end{aligned}\]

Worked example 9 — An unknown probability from a variance

A discrete score \(X\) takes values 0, 1 and 2 with probabilities \(2p\), \(1-3p\) and \(p\).

1. Find the range of \(p\) for which this is a valid probability distribution.

2. Given \(\Var(X)=\frac{14}{25}\), find \(p\) and the probability distribution of \(X\).

Solution

1. The total is \(2p+(1-3p)+p=1\) for every \(p\). The probabilities must be non-negative: \[\begin{aligned} p\geq0&\quad\Longrightarrow\quad 2p\geq0,\\ 1-3p\geq0&\quad\Longrightarrow\quad p\leq\frac13. \end{aligned}\] Therefore \(\boxed{0\leq p\leq\frac13}\). These probabilities sum to 1, so non-negativity also ensures that each is at most 1.

2. Calculate the mean and the expected value of \(X^2\): \[\begin{aligned}\E(X)&=0(2p)\\&\quad{}+1(1-3p)\\&\quad{}+2p\\&=1-p,\\\E(X^2)&=0^2(2p)\\&\quad{}+1^2(1-3p)\\&\quad{}+2^2p\\&=1+p.\end{aligned}\] Use the given variance: \[\begin{gathered}\frac{14}{25}=1+p-(1-p)^2,\\\frac{14}{25}=3p-p^2,\\25p^2-75p+14=0,\\ (5p-1)(5p-14)=0.\end{gathered}\] The roots are \(p=\frac15\) and \(p=\frac{14}{5}\). Only \(\frac15\) lies in the allowed range, so \(\boxed{p=\frac15}\).

   \(x\)	\(0\)	\(1\)	\(2\)
   \(\Prob(X=x)\)	\(\frac25\)	\(\frac25\)	\(\frac15\)

Worked example 10 — Convert a token count to a payout

For the payout \(Y=2X+3\) above, we already know \(\E(X)=\frac52\) and \(\Var(X)=\frac{13}{4}\). Find:

1. the mean payout;

2. the variance of the payout;

3. the standard deviation of the payout.

Solution

1. \(\E(Y)=2(\frac52)+3=\boxed{8\,\mathrm{euros}}\).

2. \(\Var(Y)=2^2(\frac{13}{4})=\boxed{13\,\mathrm{euros}^2}\).

3. \(\sigma_Y=\sqrt{13}=\boxed{\sqrt{13}\,\mathrm{euros}}\).

Check the mean using the transformed distribution: \[\begin{aligned}&3\left(\frac15\right)+7\left(\frac12\right)\\&\quad+13\left(\frac3{10}\right)\\&=8\,\mathrm{euros}.\end{aligned}\]

Worked example 11 — A negative multiplier

For the same envelope count, define a score \(Z=10-2X\) points. We have \(\E(X)=\frac52\) and \(\Var(X)=\frac{13}{4}\). Find:

1. \(\E(Z)\);

2. \(\Var(Z)\);

3. the standard deviation of \(Z\).

Solution

Here \(a=-2\) and \(b=10\).

1. \[\begin{aligned}\E(Z)&=(-2)(\frac52)+10\\&=\boxed{5\,\mathrm{points}}.\end{aligned}\]

2. \[\begin{aligned}\Var(Z)&=(-2)^2(\frac{13}{4})\\&=\boxed{13\,\mathrm{points}^2}.\end{aligned}\]

3. \[\begin{aligned}\sigma_Z&=|-2|(\frac{\sqrt{13}}2)\\&=\boxed{\sqrt{13}\,\mathrm{points}}.\end{aligned}\]

The outcomes \(x=0,2,5\) give \(z=10,6,0\): the negative multiplier reverses their order. Their probabilities remain \(\frac15,\frac12,\frac3{10}\) respectively. If you reorder the values, reorder their probabilities too.

Worked example 12 — Find the constants in a transformation

A random score \(X\) has mean 3 points and variance 4 points\(^2\). The adjusted score \(Y=aX+b\) has mean 8 points and standard deviation 6 points. Given \(a>0\):

1. find \(a\);

2. find \(b\).

Solution

1. First convert the standard deviation to a variance: \(\Var(Y)=6^2=36\). \[36=a^2(4)\quad\Longrightarrow\quad a^2=9.\] The condition \(a>0\) selects \(\boxed{a=3}\).

2. Use the mean equation: \[\begin{aligned}8&=a(3)+b\\&=3(3)+b\\&\Longrightarrow\boxed{b=-1}.\end{aligned}\]

Worked example 13 — Set a fair fee and interpret a loss

An envelope contains \(X=0,2\) or 5 tokens with probabilities \(\frac15,\frac12,\frac3{10}\). Each token pays 2 euros. Define \(R=2X\) as the payout in euros and \(G=R-c\) as the net gain after paying \(c\) euros. We already know \(\E(X)=\frac52\) and \(\Var(X)=\frac{13}{4}\).

1. Find the fair entry fee.

2. If the fee is 6 euros, find and interpret \(\E(G)\).

3. At the fair fee, find the probability of making a profit.

4. Find \(\Var(G)\) for any fixed fee \(c\).

Solution

1. \(\E(R)=2\E(X)=2(\frac52)=5\) euros. A fair game has \(c=\E(R)\), so the fair fee is \(\boxed{5\,\mathrm{euros}}\).

2. \[\begin{aligned}\E(G)&=\E(R)-6\\&=5-6\\&=\boxed{-1\,\mathrm{euro}}.\end{aligned}\] This is an average loss of 1 euro per play over many repetitions.

3. At \(c=5\), \(G=2X-5\) takes values \(-5,-1,5\) euros. Only \(X=5\) produces a positive net gain: \[\Prob(G>0)=\Prob(X=5)=\boxed{\frac3{10}}.\]

4. \[\begin{aligned}\Var(G)&=\Var(2X-c)\\&=2^2\Var(X)\\&=4(\frac{13}{4})\\&=\boxed{13\,\mathrm{euros}^2}.\end{aligned}\] Subtracting a fixed fee does not change the variance.

Worked example 14 — Find an unknown jackpot

In a different game, let \(R\) be the payout in euros: 0, 10 or \(J\), with probabilities \(\frac45,\frac3{20},\frac1{20}\). The fixed fee is 4 euros; define the net gain \(G=R-4\).

1. Find the jackpot \(J\) that makes the game fair.

2. For that jackpot, find the probability that the player makes a profit.

Solution

1. A fair game has \(\E(G)=0\), so \[\begin{aligned}&0(\tfrac45)+10(\tfrac3{20})\\&\quad+J(\tfrac1{20})-4=0.\end{aligned}\] Thus \(\frac32+\frac{J}{20}=4\), giving \(\boxed{J=50\,\mathrm{euros}}\).

2. The net gains are \(-4,6,46\) euros. Both payouts of 10 and 50 euros produce a profit: \[\Prob(G>0)=\frac3{20}+\frac1{20}=\boxed{\frac15}.\]

A fair game need not have a 50% chance of profit: occasional large gains can balance frequent losses.