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Expected value and variability · Tutorial 335 of 1000

Effect of Spread on Risk

Compare equal-mean distributions using standard deviation and context-specific probabilities to see how variability can affect risk.

Intermediate 9 min read

What You'll Learn

  • Compare the variability of two distributions with equal means using their standard deviations.
  • Calculate and interpret standard deviations in the original units of the random variable.
  • Connect greater spread to a higher chance of outcomes far from the mean.
  • Use a clearly defined unfavorable outcome to compare risk in context.
  • Explain why standard deviation alone does not determine the probability of a particular loss or shortfall.

Same Average, Different Risk

In Mean of a Random Variable with a Missing Probability, you completed a distribution before calculating its mean and standard deviation. Now consider what happens when two complete distributions have the same mean but different standard deviations. Their long-run averages match, but their outcomes can vary by very different amounts.

The mean describes a probability-weighted center. Standard deviation describes the typical distance of values from that center, in the original units. As explained in Interpreting Standard Deviation of a Random Variable, it is a measure of variability, not a prediction of the next outcome. When unusually low or high outcomes matter, more variability can create more risk—but what counts as “risk” depends on the situation.

Key idea: If two options have the same mean, the one with the larger standard deviation has outcomes that typically lie farther from the shared mean. That greater spread may mean more risk, but a specific risk comparison should identify which outcomes are unfavorable.

For an investment, a return far below the expected return may be a concern. For a cost, a value far above the expected cost may be the unfavorable outcome. Standard deviation treats distances above and below the mean alike; it does not by itself tell you the probability of crossing a particular loss or cost threshold.

Comparing Spread and Risk

To compare distributions fairly, use the same time period and units, and compare their means first. When the means match, standard deviations help show how much the outcomes differ in typical distance from that shared center. A larger standard deviation signals greater overall variability.

Recall the formulas from Variance of a Discrete Random Variable and The Standard Deviation of a Discrete Random Variable. For possible values \(x\), variance is the probability-weighted average of squared distances from the mean, and standard deviation is the square root of variance:

$$ \sigma_X^2=\sum (x-\mu_X)^2P(X=x) \qquad\text{and}\qquad \sigma_X=\sqrt{\sigma_X^2}. $$

A useful comparison has three parts: state whether the means are equal, compare the standard deviations, and explain what that spread means in context. If the question asks about a particular unfavorable result, calculate or compare its probability directly as well.

1
Define what varies.
State what each random variable measures, including the time period and units.
2
Compare the centers.
Calculate or identify each mean. Confirm whether the distributions have the same expected value.
3
Compare the spread.
Find each standard deviation and describe the difference using the variable’s original units.
4
Connect spread to the stated risk.
Identify what outcome is unfavorable. Use the standard deviations to discuss overall variability and probabilities to compare a particular threshold.

Worked Example: Two Investment Plans

Worked Example: Two Investment Plans

A fictional model describes the one-year return \(X\), in dollars, for each of two investment plans. Plan A returns $100 with certainty. Plan B returns $80 with probability 0.25, $100 with probability 0.50, and $120 with probability 0.25. Compare their means and standard deviations. If a return below $90 is considered unfavorable, compare the plans’ risk of that outcome.

Plan A return, dollarsProbabilityPlan B return, dollarsProbability
1001.00800.25
1000.50
1200.25

State. Let \(X_A\) and \(X_B\) be the one-year returns, in dollars, for Plans A and B. We will compare their expected returns, their variability, and the probability of a return below $90.

Plan. Each distribution is complete: its probabilities are between 0 and 1 and sum to 1. Calculate each mean and standard deviation from its value-probability pairs. Then identify which outcomes satisfy the unfavorable condition \(X<90\).

Do. Plan A always returns $100, so its mean is $100. Every possible return is exactly at the mean, so its variance and standard deviation are both 0.

For Plan B, calculate the mean:

$$ \begin{aligned} \mu_{X_B} &=(80)(0.25)+(100)(0.50)+(120)(0.25)\\ &=20+50+30\\ &=100\text{ dollars}. \end{aligned} $$

The means are equal. Plan B’s variance, using weighted squared deviations, is:

$$ \begin{aligned} \sigma_{X_B}^2 &=(80-100)^2(0.25)+(100-100)^2(0.50)+(120-100)^2(0.25)\\ &=(400)(0.25)+0+(400)(0.25)\\ &=200\text{ dollars}^2,\\ \sigma_{X_B}&=\sqrt{200}\approx14.1421\text{ dollars}. \end{aligned} $$

Check the variance using \(\sum x^2P(X=x)-\mu_X^2\):

$$ \begin{aligned} \sigma_{X_B}^2 &=(80^2)(0.25)+(100^2)(0.50)+(120^2)(0.25)-(100)^2\\ &=1600+5000+3600-10000\\ &=200\text{ dollars}^2. \end{aligned} $$

The second calculation agrees, and the square root is about $14.1421. For the unfavorable outcome, Plan A has no possible return below $90, so \(P(X_A<90)=0\). Under Plan B, only the $80 return is below $90, so \(P(X_B<90)=0.25\), or 25%.

Conclude. Both plans have an expected one-year return of $100, but Plan B’s returns typically lie farther from that mean: its standard deviation is about $14.14, compared with $0 for Plan A. In this model, Plan B has a 0.25 (25%) chance of returning less than $90, while Plan A has no chance of doing so. Plan B is more variable and has more risk of the stated unfavorable outcome.

Worked Example: Comparing Possible Repair Costs

Worked Example: Comparing Possible Repair Costs

A fictional service company is comparing two plans for the repair cost \(C\), in dollars, for one device during a year. Plan P has costs of $40, $50, and $60, each with probability 0.25, 0.50, and 0.25, respectively. Plan Q has costs of $30, $50, and $70, with those same probabilities. Compare the means and standard deviations. Treat a cost of at least $65 as an unfavorable high cost.

State. Let \(C_P\) and \(C_Q\) be the yearly repair costs, in dollars, for Plans P and Q. We need to compare expected costs, variability, and the chance that the cost is at least $65.

Plan. Both distributions are valid: each listed probability is between 0 and 1, and \(0.25+0.50+0.25=1\). Calculate each mean and standard deviation, then add the probabilities of outcomes at or above $65.

Do. The means are:

$$ \begin{aligned} \mu_{C_P}&=(40)(0.25)+(50)(0.50)+(60)(0.25)=10+25+15=50\text{ dollars},\\ \mu_{C_Q}&=(30)(0.25)+(50)(0.50)+(70)(0.25)=7.5+25+17.5=50\text{ dollars}. \end{aligned} $$

For Plan P, the variance and standard deviation are:

$$ \begin{aligned} \sigma_{C_P}^2 &=(40-50)^2(0.25)+(50-50)^2(0.50)+(60-50)^2(0.25)\\ &=100(0.25)+0+100(0.25)=50\text{ dollars}^2,\\ \sigma_{C_P}&=\sqrt{50}\approx7.0711\text{ dollars}. \end{aligned} $$

For Plan Q:

$$ \begin{aligned} \sigma_{C_Q}^2 &=(30-50)^2(0.25)+(50-50)^2(0.50)+(70-50)^2(0.25)\\ &=400(0.25)+0+400(0.25)=200\text{ dollars}^2,\\ \sigma_{C_Q}&=\sqrt{200}\approx14.1421\text{ dollars}. \end{aligned} $$

As a computational check, Plan P has \(\sum c^2P(C_P=c)=2550\), so its variance is \(2550-50^2=50\). Plan Q has \(\sum c^2P(C_Q=c)=2700\), so its variance is \(2700-50^2=200\). These match the weighted squared-deviation calculations. For the high-cost event \(C\geq65\), only the $70 outcome in Plan Q qualifies. Thus \(P(C_Q\geq65)=0.25\), while \(P(C_P\geq65)=0\).

Conclude. Both plans have an expected yearly repair cost of $50, but Plan Q has twice the standard deviation of Plan P, about $14.14 rather than $7.07. Plan Q also has a 0.25 (25%) chance of a cost of at least $65, compared with no chance under Plan P. If avoiding high costs is important, Plan P has less variability and less risk of the specified high-cost outcome.

Spread Does Not Answer Every Risk Question

Standard deviation describes overall spread around the mean; it does not tell you exactly how probability is distributed on either side of the mean. Two distributions with different shapes can have different chances of a particular unfavorable outcome, even when their standard deviations suggest different overall variability. Therefore, use standard deviation to compare spread, and calculate a probability when the risk question names a cutoff.

Worked Example: A Rare Severe Loss and a Frequent Small Loss

Two fictional investment models have the same mean one-year return of $100. Investment A returns $99 or $101 with probability 0.50 each. Investment B returns $80 with probability 0.10 and \(\frac{920}{9}\) dollars (about $102.22) with probability 0.90. Compare their standard deviations and the probabilities of a return below $100.

State. Let \(X_A\) and \(X_B\) be the one-year returns, in dollars. We will compare overall variability with the specific risk event \(X<100\).

Plan. The probabilities for each investment are between 0 and 1 and total 1. Calculate the mean and standard deviation for each distribution, then identify the outcomes below $100.

Do. Investment A’s mean is \((99)(0.50)+(101)(0.50)=100\) dollars. Its variance and standard deviation are:

$$ \begin{aligned} \sigma_{X_A}^2 &=(99-100)^2(0.50)+(101-100)^2(0.50)\\ &=1(0.50)+1(0.50)=1\text{ dollars}^2,\\ \sigma_{X_A}&=\sqrt{1}=1\text{ dollar}. \end{aligned} $$

For Investment B, its mean is:

$$ \begin{aligned} \mu_{X_B} &=(80)(0.10)+\left(\frac{920}{9}\right)(0.90)\\ &=8+92\\ &=100\text{ dollars}. \end{aligned} $$

The value \(\frac{920}{9}\) is \(102\frac{2}{9}\), so its distance above the mean is \(\frac{20}{9}\) dollars. Investment B’s variance and standard deviation are:

$$ \begin{aligned} \sigma_{X_B}^2 &=(80-100)^2(0.10)+\left(\frac{920}{9}-100\right)^2(0.90)\\ &=400(0.10)+\left(\frac{20}{9}\right)^2(0.90)\\ &=40+\frac{40}{9}=\frac{400}{9}\text{ dollars}^2,\\ \sigma_{X_B}&=\sqrt{\frac{400}{9}}=\frac{20}{3}\approx6.6667\text{ dollars}. \end{aligned} $$

The computational check gives \(\sum x^2P(X_A=x)=10001\), so Investment A’s variance is \(10001-100^2=1\). For Investment B, the second moment is \(80^2(0.10)+(\frac{920}{9})^2(0.90)=10000+\frac{400}{9}\), giving variance \(\frac{400}{9}\) after subtracting \(100^2\). For the specified event, Investment A returns less than $100 only when it returns $99, so \(P(X_A<100)=0.50\). Investment B returns less than $100 only when it returns $80, so \(P(X_B<100)=0.10\).

Conclude. Investment B has a larger standard deviation, about $6.67 compared with $1 for Investment A, because its possible outcomes are farther from the mean overall. Yet Investment A has the greater probability of any return below $100: 0.50 compared with 0.10. Investment B’s below-$100 outcome is less frequent but more severe. A decision-maker concerned about a specific threshold should consider both its probability and how far the outcome falls below that threshold, not standard deviation alone.

Common Mistakes and AP Exam Tips

  • Concluding that equal means imply equal risk. Equal means describe equal probability-weighted averages, not equal spreads. Compare the standard deviations and explain their meaning in context.
  • Calling the standard deviation a guaranteed distance. A standard deviation describes typical distance from the mean; it does not say that every outcome is within one standard deviation.
  • Confusing a probability with a percent. A probability of 0.25 is equivalent to 25%, not 0.25%. Keep the decimal or percent notation consistent and state which one you mean.
  • Assuming greater standard deviation always means greater probability of a particular loss. Standard deviation measures overall spread on both sides of the mean. For an event such as a return below $90, calculate that event’s probability directly.
  • Forgetting that unfavorable depends on context. A high return may be desirable for an investor, while a high cost may be undesirable. Define the random variable and the relevant unfavorable event before interpreting risk.
  • Comparing unlike quantities. Compare distributions over the same time period and in the same units. A standard deviation in dollars cannot be directly compared with one in days as if the sizes represented the same variability.

A full-credit comparison states the shared mean, gives both standard deviations with units, and explains which distribution has greater variability. If a particular risk threshold is mentioned, write the event in probability notation, identify the qualifying outcomes, and interpret the probability in context.

Key Takeaway

When two distributions share a mean, the distribution with the larger standard deviation has more overall variability: its outcomes typically lie farther from the common center. Greater variability can increase risk, but risk is tied to the outcomes that matter in context. A specific chance of a loss, low return, or high cost should be assessed with its own probability.

Key takeaway: Compare means to assess expected outcomes and standard deviations to assess overall spread. To compare a particular unfavorable outcome, define its threshold and calculate its probability; standard deviation alone does not determine that probability.

Check Your Understanding

Use the following fictional distribution comparisons to practice connecting equal means, spread, and risk.

  1. Option A returns $60 or $100 with probabilities 0.50 and 0.50. Option B returns $40, $80, or $120 with probabilities 0.25, 0.50, and 0.25. Find both means and standard deviations, showing the variance calculations.
  2. For the two options in question 1, compare the probability of a return below $70. Which option has greater variability, and which has a higher probability of that specific unfavorable outcome?
  3. A service plan has a mean yearly cost of $300 and standard deviation of $20; another has the same mean and standard deviation of $45. What does the larger standard deviation indicate, and what does it not establish about the chance that cost exceeds $350?
  4. In a complete comparison of two equal-mean distributions, what should an AP response say about the means, standard deviations, units, and any specified risk threshold?

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