ML Interview Q Series: Unpacking the MGF Condition (M_X(t) = e^t M_X(-t)): Symmetry and Expectation.
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Suppose that the moment-generating function (M_{X}(t)) of the continuous random variable (X) satisfies
Short Compact solution
By rewriting this condition as
(M_{X}(t) = \mathrm{E}[,e^{,tX},] = e^{,t},\mathrm{E}[,e^{-,tX},] = e^{,t},\mathrm{E}[,e^{-,t,(,1 - X,)},] = \mathrm{E}[,e^{,t,(,1 - X,)},] = M_{1 - X}(t),)
we see that (X) has the same distribution as (1 - X). This immediately gives
(\mathrm{E}[X] = \mathrm{E}[,1 - X,] \quad \Longrightarrow \quad \mathrm{E}[X] = \tfrac{1}{2}.)
However, the condition (M_{X}(t) = e^{t}M_{X}(-t)) does not uniquely determine the density of (X). For instance, both the uniform distribution on ((0,1)) and the beta distribution with pdf (6,x,(1 - x)) on ((0,1)) satisfy this property.
Comprehensive Explanation
Deriving (E(X))
First, observe the moment-generating function (MGF) relationship:
Interpreting (M_{X}(t)) as (\mathrm{E}[,e^{tX},]), we write:
[ \huge \mathrm{E}[,e^{tX},] ;=; e^{t},\mathrm{E}[,e^{-tX},] ;=; e^{t},\mathrm{E}[,e^{-t,(1 - X)},] ;=; \mathrm{E}[,e^{t,(1 - X)},]. ]
Thus,
Since the MGF uniquely determines a distribution (provided it exists in a neighborhood of (t=0)), we see that the random variable (X) has the same distribution as (1 - X). Two random variables that share the same distribution will have identical means. Therefore,
[ \huge \mathrm{E}[X] ;=; \mathrm{E}[,1 - X,] ;=; 1 - \mathrm{E}[X] \quad\Longrightarrow\quad \mathrm{E}[X] ;=; \tfrac{1}{2}. ]
Does the condition uniquely determine the density?
Even though the MGF typically characterizes a distribution completely if it exists for all (t) in an open interval containing zero, the symmetry-like condition (M_{X}(t) = e^{t} M_{X}(-t)) alone is not strong enough to rule out multiple distributions. What it guarantees is that (X) and (1 - X) share the same distribution, but it does not specify exactly which distribution it is—only that it must be symmetric around 1/2 in some sense.
For example:
Uniform(0,1) distribution. In this case, (X) is symmetric about 1/2 in the sense that (X) and (1-X) follow the same uniform(0,1) distribution. It satisfies the given MGF condition.
Beta(2,2) distribution (with pdf (6,x,(1-x)) over ((0,1))). This distribution is also symmetric around 1/2 and satisfies the same MGF relation.
Hence, the property (M_{X}(t) = e^{t},M_{X}(-t)) forces (\mathrm{E}[X]=1/2) but does not fully determine the distribution of (X).
Follow-up Questions
1. Why does having the same MGF typically imply the same distribution, yet here we still have multiple distributions?
When we say “having the same MGF” in a general sense, we mean two distinct random variables share the entire MGF function for all real (t) in some interval around 0. Under standard conditions (the MGF should be finite in some open interval around 0), that determines the distribution uniquely.
However, in our scenario, we are dealing with the functional equation (M_{X}(t) = e^{t} M_{X}(-t)). This functional equation holds for multiple specific forms of (M_{X}(\cdot)). All those MGFs satisfying it will make (X) and (1 - X) identically distributed but do not necessarily reduce to just one unique functional form. The existence of more than one valid MGF solution means more than one distribution can fit that property.
2. How can we verify that a specific distribution satisfies (X \stackrel{d}{=} 1 - X)?
To verify (X) is identically distributed as (1 - X), you can check if the pdf (or cumulative distribution function) is symmetric around 1/2 in the interval ((0,1)). Specifically:
For a continuous distribution on ((0,1)), having (f_{X}(x) = f_{X}(1-x)) for all (x \in (0,1)) is one equivalent way to say (X) has the same distribution as (1 - X).
Alternatively, you can compute the MGF (or characteristic function) and check if (M_{X}(t) = M_{1-X}(t)).
3. Could we have discrete distributions with the same property?
Yes, the same logic applies to discrete variables or even mixed variables. If (X) takes values in ([0,1]) in a discrete manner but satisfies (X\stackrel{d}{=}1-X), it would similarly mean the probability mass function is symmetric around 1/2. An example would be a random variable that takes values ({0,1}) each with probability 1/2.
4. How can we construct more distributions satisfying (X \stackrel{d}{=} 1 - X)?
Any distribution supported on ((0,1)) (or even on ([-a, a]) but then shifted and scaled to the interval ((0,1))) that is symmetric about the midpoint 1/2 in its support can satisfy (X\stackrel{d}{=}1-X). We can do this by defining a pdf (f(x)) such that (f(x)=f(1-x)) for (x) in ((0,1)). Then the condition on MGFs holds. Examples:
Uniform(0,1).
Beta((\alpha,\alpha)) for any (\alpha>0).
More complicated symmetric pdfs that are carefully normalized over (0,1).
5. Is there a direct relationship between the characteristic function and this MGF property?
Yes. The characteristic function (\phi_{X}(t)=\mathrm{E}[e^{,i,tX}]) has a parallel symmetry property if (X) and (1 - X) share the same distribution. Specifically, (\phi_{X}(t)) would satisfy (\phi_{X}(t) = e^{,i,t},\phi_{X}(-,t)). The moment-generating function is simply the characteristic function evaluated at purely real arguments rather than purely imaginary arguments. The distribution symmetry argument applies equally well to both transforms under the standard existence assumptions.
6. Could this property hold if the support of (X) were unbounded?
In theory, if (X) has support on all real numbers, then the condition (X\stackrel{d}{=}1-X) would imply that the distribution is symmetric about 1/2 in a sense that goes beyond the ((0,1)) interval. However, for many unbounded distributions, we might run into issues with the existence of the MGF across the entire real line or even in a symmetric interval around 0. In practice, many well-known unbounded distributions (e.g., normal, exponential) do not satisfy (X\stackrel{d}{=}1-X). The property can hold only if the distribution is truly mirrored around 1/2. It is more straightforwardly recognized in bounded distributions on ((0,1)) or discrete analogs on {0,1}, etc.
7. Can we numerically check this property for a given dataset?
In a practical setting with real data, suppose you suspect your data might follow a distribution that is symmetric about 1/2. You could:
Shift the data by subtracting 1/2 from each observation to see if it is symmetric about 0 (e.g., check if the mean is near 0, or check if the distribution is symmetrical around 0).
Or directly check whether the empirical distribution function (\hat{F}(x)) satisfies (\hat{F}(x)\approx 1-\hat{F}(1-x)) for all (x) in the observed range.
However, verifying an MGF relationship empirically is rarely done in raw form. One usually checks simpler symmetry or performs distribution-fitting tests.
Overall, the key takeaway is that the functional equation ensures (\mathrm{E}[X]=1/2) but does not lock us into a single distribution. It only enforces the symmetry (X \stackrel{d}{=} 1 - X), and multiple densities satisfy this.
Below are additional follow-up questions
1. How does the given MGF property affect higher-order moments, such as Var(X) or E(X^2)?
Because we know E(X) = 1/2, one natural follow-up is to see what this condition implies about higher-order moments. In principle, the MGF can provide all moments by successive derivatives at t = 0. However, the relation M_X(t) = e^t M_X(-t) by itself does not force a single unique variance or higher moments. Instead, each valid distribution satisfying X = 1 – X in distribution can come with different second moments.
A subtle real-world pitfall here is to assume that if E(X) is pinned down, then Var(X) must also be pinned down. That is not the case because multiple densities with different shapes can all be symmetric about 1/2. For instance:
A uniform(0,1) random variable has variance 1/12.
A Beta(α, α) random variable, also symmetric, has variance α(α + 1) / [ (2α + 1)(2α)^2 ], which will vary with α.
Hence, while the condition yields E(X) = 1/2, the second moment and higher moments remain distribution-dependent.
2. Are there continuous distributions outside the interval (0,1) that satisfy X = 1 – X in distribution?
Yes. The property that X is distributed as 1 – X translates to a reflection symmetry about 1/2. However, you can shift and scale any random variable symmetric about a midpoint and transform it onto (0,1). For instance, consider a variable Y that is symmetric about 0 (e.g., a standard normal). Define X = 1/2 + (Y / (2c)) for some constant c chosen such that X remains in (0,1) with high probability (or strictly, with a truncated normal). Then X could, in principle, be symmetric around 1/2 within a bounded interval.
A subtle pitfall arises if we consider unbounded supports without truncation. For a truly unbounded distribution, “X = 1 – X in distribution” implies that the distribution’s center of symmetry is 1/2, but actually enforcing that on the real line can be tricky. It typically forces the distribution to be bounded or else have infinite tails that mirror each other in a way not commonly seen in typical unbounded families (like normal or exponential).
3. Could a mixture of distributions satisfy this property?
Absolutely. If each component of a mixture satisfies the symmetry condition (meaning each component distribution is such that X = 1 – X in distribution), then any convex combination of those components retains the property. For instance, if X_1 is uniform(0,1) and X_2 is Beta(2,2), each separately satisfies the reflection symmetry. A mixture X with probability p coming from X_1 and probability (1 – p) coming from X_2 will also satisfy X = 1 – X in distribution.
A subtlety arises when one attempts to mix two distributions that both individually have E(X) = 1/2, but differently shaped pdfs. One might incorrectly assume the mixture changes the mean. In fact, the mean is still 1/2, but the final shape (and thus higher-order moments) can change significantly. This is an important consideration in real-world modeling, where data might come from multiple sources yet still exhibit overall symmetry around 1/2.
4. Is it possible for a degenerate (deterministic) random variable to satisfy M_X(t) = e^t M_X(-t)?
A degenerate random variable at a constant c has an MGF of exp(c t). If it were to satisfy M_X(t) = e^t M_X(-t), we would need exp(c t) = e^t exp(–c t). That simplifies to exp(c t) = exp(t – c t), or c t = t – c t. Then 2 c t = t, so c = 1/2 for all t ≠ 0. This yields exactly X = 1/2 with probability 1. That trivial scenario does indeed satisfy X = 1 – X in distribution because 1/2 = 1 – 1/2. It is a trivial distribution but a valid edge case.
5. Could this condition hold if the MGF does not exist for all t in some open interval containing 0?
MGFs need to exist (be finite) in at least a neighborhood around t = 0 to uniquely define a distribution via the usual theorems. However, certain heavy-tailed distributions may fail to have an MGF that converges except at t = 0 (e.g., Cauchy distribution). For such distributions, the statement M_X(t) = e^t M_X(-t) for all t might not even make sense. Therefore, the requirement that the MGF be well-defined over some open interval containing 0 is crucial to employ the typical “MGF characterizes distribution” argument.
A real-world pitfall is to see the functional equation M_X(t) = e^t M_X(-t) in an interview or textbook context and assume it always applies if the distribution is “nice enough.” But not all random variables are “nice enough” to have a well-defined MGF beyond t = 0. One must verify the domain of definition for the MGF to ensure the argument about E(X) = 1/2 holds.
6. Could this property be extended to random vectors in higher dimensions?
One might look for an analog in multiple dimensions: M_X(t) = e^(t^T * c) M_X(–t) for some vector c, implying that X is distributed the same as c – X. A direct analog in 1D is c = 1, giving X = 1 – X. In higher dimensions, c would be a constant vector representing a “center of symmetry.” If we define c = (1, 1, ..., 1), then X is distributed as c – X, meaning the distribution is centrally symmetric about the point (1/2, 1/2, ..., 1/2).
One pitfall: while the concept is straightforward, verifying or imposing MGF conditions in multiple dimensions can be more complex. MGFs are now functions of a vector argument, and a symmetrical condition might require more careful bounding of the domain of existence. This can be relevant in real-world problems involving symmetrical shapes in a hypercube, but is rarely encountered in standard univariate textbook examples.
7. If empirical data show a sample mean far from 1/2, does that disprove X = 1 – X in distribution?
In a real dataset, sampling variability is expected. Even if the true distribution has mean 1/2, a finite sample might show a sample mean that deviates somewhat from 1/2. However, if the deviation is extremely large (beyond what you would reasonably expect from sampling error), that is evidence the distribution might not have that reflection symmetry.
A subtle pitfall is conflating sample mean with the actual theoretical mean. With limited samples, you might incorrectly conclude the distribution is not symmetric simply because the empirical mean doesn’t exactly match 1/2. Standard statistical testing for symmetry (for example, comparing f̂(x) and f̂(1 – x) in an empirical sense) can be more direct. Small sample sizes can mislead such tests, so domain-specific knowledge and caution are necessary before drawing definitive conclusions.
8. In practical modeling, why might we rarely see this property explicitly used?
Most real-world phenomena do not naturally fall into the pattern that X = 1 – X in distribution, unless the domain is specifically restricted to [0,1] or the data are artificially normalized to that interval. Additionally, many real-world processes lack such perfect symmetry around 1/2. Even in scenarios where variables are constrained to [0,1], one often encounters skewed distributions (e.g., success probabilities in Bernoulli trials) rather than symmetrical Beta(α,α) distributions.
A pitfall in data science is forcing a symmetrical model on data that exhibit asymmetry. Doing so can degrade model performance and misrepresent uncertainty. While having E(X) = 1/2 may simplify certain analyses, it is rarely an accurate reflection of real data unless the phenomenon genuinely suggests that symmetrical structure.