Direct conclusion
For a perfectly symmetrical distribution which relationship is always true? The always-true relationship (assuming the mean exists) is \[ \text{mean}=\text{median}. \] The equality \(\text{mean}=\text{median}=\text{mode}\) is not guaranteed unless the distribution is unimodal and symmetric.
What “perfectly symmetrical” means
A distribution is perfectly symmetric about a point \(c\) if the left and right sides mirror each other. In density terms (continuous case), symmetry means:
\[ f(c+a)=f(c-a)\quad \text{for all }a \text{ where both sides are defined.} \]
In probability terms, the deviations from \(c\) are balanced: \[ P(X \le c-a)=P(X \ge c+a)\quad \text{for all }a \ge 0. \]
Why the median equals the center of symmetry
The median \(m\) is any value satisfying \(P(X \le m)\ge 0.5\) and \(P(X \ge m)\ge 0.5\). For a distribution symmetric about \(c\), the mirror-balance implies exactly half the probability lies on each side of \(c\), so:
\[ P(X \le c)=0.5 \quad \text{(for continuous symmetric distributions)}. \]
Therefore, the median is \(c\) (or one of the medians equals \(c\) in discrete/flat cases).
Why the mean equals the center of symmetry (when the mean exists)
If the mean exists, symmetry forces positive and negative deviations from \(c\) to cancel in expectation. Define \(Y=X-c\). Symmetry about \(c\) is equivalent to \(Y\) being symmetric about \(0\), meaning \(Y\) and \(-Y\) have the same distribution. Then:
\[ \mathbb{E}(Y)=\mathbb{E}(-Y)=-\mathbb{E}(Y)\ \Rightarrow\ \mathbb{E}(Y)=0. \]
\[ \mathbb{E}(X-c)=0\ \Rightarrow\ \mathbb{E}(X)=c. \]
Since the median also equals \(c\), the always-true relationship is \(\text{mean}=\text{median}\).
What about the mode?
The mode is the location of a highest peak (maximum density or maximum probability). Symmetry alone does not force a unique peak at the center:
- If the distribution is unimodal and symmetric, the peak is at the center, so \(\text{mode}=\text{median}=\text{mean}=c\).
- If the distribution is bimodal and symmetric, there can be two modes equidistant from \(c\), so \(\text{mode}\) need not equal the mean/median.
| Distribution type | Always true | Not always true |
|---|---|---|
| Symmetric about \(c\) (mean exists) | \(\text{mean}=\text{median}=c\) | \(\text{mode}=c\) |
| Unimodal and symmetric about \(c\) | \(\text{mean}=\text{median}=\text{mode}=c\) | (No exception under these conditions) |
| Symmetric with two equal peaks | \(\text{mean}=\text{median}=c\) | \(\text{mode}=c\) (often false) |
Visualization: a symmetric unimodal distribution
Quick check with a symmetric but bimodal example
Consider a distribution that puts probability \(0.5\) at \(-1\) and probability \(0.5\) at \(+1\). It is perfectly symmetric about \(0\). Then:
\[ \text{mean}=\mathbb{E}(X)=0.5(-1)+0.5(1)=0,\quad \text{median}=0. \]
However, the modes are \(-1\) and \(+1\), not \(0\). This shows why \(\text{mode}=\text{mean}\) is not an always-true statement for symmetry alone.