How Does Expected Value Change with the Probability Measure?

The expected value of a random variable is not determined by the random variable alone. It also depends on the probability measure used to weight the possible outcomes. Thus, the same function on the same sample space can have different expected values under different probability measures.

The standard meaning of this question is: given a random variable and two probability measures, how do you define and compare its expectations? The answer is to integrate the random variable with respect to each measure separately.

Expectation is an integral with respect to a measure

Let $(\Omega,\mathcal{F})$ be a measurable space, and let $X: \Omega\to\mathbb{R}$ be a measurable random variable. If $P$ is a probability measure on that space, then

$$\mathbb{E}_P[X]=\int_\Omega X(\omega)\,P(d\omega),$$

provided the expectation exists. For a real-valued $X$, a finite expectation is ensured by $\mathbb{E}_P[|X|]<\infty$. For nonnegative $X$, the integral is defined even if it is infinite.

If $Q$ is another probability measure on the same measurable space, then its expectation is

$$\mathbb{E}_Q[X]=\int_\Omega X(\omega)\,Q(d\omega).$$

The function $X$ has not changed, but the measure assigning probabilities to outcomes has. There is no general reason for these two integrals to be equal.

A finite example

Take $\Omega=\{0,1,2\}$ and let $X(\omega)=\omega$. Define two probability measures by

Outcome $\omega$	$P(\{\omega\})$	$Q(\{\omega\})$
0	$1/2$	$1/6$
1	$1/3$	$1/3$
2	$1/6$	$1/2$

Each column of probabilities sums to 1. Under $P$,

$$\mathbb{E}_P[X]=0\cdot\frac12+1\cdot\frac13+2\cdot\frac16=\frac23.$$

Under $Q$,

$$\mathbb{E}_Q[X]=0\cdot\frac16+1\cdot\frac13+2\cdot\frac12=\frac43.$$

The expected value is larger under $Q$ because $Q$ gives more probability to the larger outcome 2. This illustrates the basic rule: expectation averages values using the probabilities supplied by the chosen measure.

Changing measure with a density

Sometimes one measure can be described as a reweighting of another. If $Q$ is absolutely continuous with respect to $P$, written $Q\ll P$, the Radon–Nikodym theorem gives a measurable density

$$L=\frac{dQ}{dP}$$

such that $Q(A)=\int_A L\,dP$ for measurable sets $A$. For a suitable integrable random variable $X$, the change-of-measure formula is

$$\mathbb{E}_Q[X]=\mathbb{E}_P[XL].$$

The density $L$ increases the weight of outcomes that are more likely under $Q$ than under $P$, and decreases the weight of outcomes that are less likely. Since $Q$ is a probability measure, $\mathbb{E}_P[L]=1$.

In the finite example, every outcome has positive probability under $P$, so $Q\ll P$. The density values are $L(0)=1/3$, $L(1)=1$, and $L(2)=3$. Therefore,

$$\mathbb{E}_P[XL]=0\cdot\frac12\cdot\frac13+1\cdot\frac13\cdot1+2\cdot\frac16\cdot3=\frac43,$$

which agrees with $\mathbb{E}_Q[X]$.

If $Q$ is not absolutely continuous with respect to $P$, this density formula cannot in general be used. For example, if $P$ assigns probability zero to an event that has positive probability under $Q$, then no density $dQ/dP$ exists. You can still define $\mathbb{E}_Q[X]$ directly, as long as $X$ is measurable and integrable under $Q$.

When expectations are the same

Different measures can nevertheless give the same expectation for a particular variable. In the finite example, the measures differ, but both give probability $1/3$ to outcome 1. Consequently, the indicator variable $X(\omega)=\mathbf{1}_{\{1\}}(\omega)$ has expectation $1/3$ under both.

More generally, if $X$ has the same distribution under $P$ and $Q$, then its expected value is the same under both measures whenever that expectation exists. Equal distributions are sufficient, but they are not necessary for two particular expectations to agree.

Common mistakes

• Writing $\mathbb{E}[X]$ without specifying the measure when it matters. In a setting with multiple measures, use $\mathbb{E}_P[X]$ and $\mathbb{E}_Q[X]$ to make the choice explicit.
• Changing the measure but not the probabilities in a discrete calculation. The expectation is a sum of values weighted by the probabilities under the selected measure.
• Using a density without checking absolute continuity. The formula $\mathbb{E}_Q[X]=\mathbb{E}_P[X(dQ/dP)]$ requires $Q\ll P$ and appropriate integrability.
• Assuming every expectation is finite. For an unbounded random variable, verify integrability under the measure being used.

If the phrase “different probability measures” instead means a random object whose values are themselves probability measures, that is a separate setup. One must specify a probability law on the space of measures before taking an expectation of that random object.