Maximum a posteriori estimation (MAP): Worked examples

This page is a drill. Its parent, MAP estimation, showed you the machinery: $\hat{θ}_{MAP} = ar g max_{θ} [lo g P (x ∣ θ) + lo g P (θ)]$ . Here we throw every kind of situation at that machinery so no exam question can surprise you.

First, one symbol we will use on every line:

Now a tiny reminder of the words, in plain language, each anchored to a picture below:

MAP is just "find the highest point of likelihood times prior." Everything below is that one sentence, stress-tested.

The scenario matrix

Every MAP problem falls into one of the eight cells A–H below. Read the table as "cell → what makes it tricky → which example drills it." Each worked example later is tagged with its cell letter, so you can hunt any scenario by its label. Think of A–H as a checklist: if you can do all eight, no MAP question is new to you.

Cell	Cell class	What makes it tricky	Example
A	Prior disagrees with data	prior pulls estimate one way, data pulls the other	Ex 1
B	Data-scarce vs data-rich limit	small $n$ (prior dominates) → large $n$ (data dominates)	Ex 2
C	Degenerate: zero successes ( $k = 0$ )	MLE gives $0$ ; does MAP?	Ex 3
D	Degenerate: uniform prior	MAP must collapse exactly to MLE	Ex 4
E	Mode $\neq =$ mean (asymmetric posterior)	which number to report	Ex 5
F	Regularization face (Gaussian prior = L2)	see the penalty become a prior	Ex 6
G	Real-world word problem	translate English → likelihood + prior	Ex 7
H	Exam twist: prior with $α < 1$ or $β < 1$ forcing a boundary solution	derivative never zero inside; check the edges	Ex 8

Prerequisites in reach: Bayes Theorem, Beta Distribution, Posterior Distribution, Conjugate Priors, Maximum Likelihood Estimation (MLE), Regularization in ML.

Figure — Maximum a posteriori estimation (MAP)

Figure 1 (read me first). The blue curve is the likelihood $P (x ∣ θ)$ , the pink curve is the prior $P (θ)$ , and the yellow curve is their point-by-point product, the posterior. The dashed yellow line marks its peak — the MAP estimate. This single picture is the template for Examples 1–5 and 8: those all live on this $[0, 1]$ axis, just with different blue and pink shapes. Keep it in view.

Coin-flip cells (Beta prior + Binomial data)

Whenever the data is "successes out of trials," the likelihood is Binomial and the natural prior is a Beta. Recall the Beta $(α, β)$ shape $P (θ) \propto θ^{α - 1} (1 - θ)^{β - 1}$ for $θ \in [0, 1]$ . Because Beta is the conjugate prior for the Binomial, the posterior is again Beta:

Why this mode formula, and why the edge checks? Take $lo g [θ^{a - 1} (1 - θ)^{b - 1}]$ and differentiate: the slope is $\frac{a - 1}{θ} - \frac{b - 1}{1 - θ}$ . Setting it to $0$ gives the fraction. But a stationary point only exists inside $(0, 1)$ when both $a - 1$ and $b - 1$ have the sign that makes the two terms trade off. If, say, $a - 1 \leq 0$ , the first term never pulls the slope positive, so the function only falls — its highest point is the boundary $θ = 0$ . That is why you cannot blindly plug in: the " $- 1$ "s can push you onto an edge. (The same " $- 1$ " mismatch also makes the mode differ from the mean — a distinction Ex 5 hammers home.)

Worked example Example 1 — Cell A: prior fights the data

Statement: 8 heads in 10 flips, so $x$ has $n = 10$ entries with $k = 8$ heads. Prior Beta $(2, 8)$ (you strongly believed the coin favours tails). Find $\hat{θ}_{MAP}$ and compare to MLE.

Forecast: MLE is $0.8$ . The prior mean is $2/10 = 0.2$ . Guess: MAP lands somewhere between, closer to $0.8$ since 10 flips is real evidence. Write your guess down.

Identify $k, n, α, β$ . $k = 8, n = 10, α = 2, β = 8$ . Why this step? The formula only needs these four numbers.
Check the edges, then plug in. Posterior shape numbers $a = k + α = 10$ , $b = n - k + β = 10$ ; both $> 1$ , so the interior formula is valid: $\hat{θ}_{MAP} = \frac{10 - 1}{10 + 10 - 2} = \frac{9}{18} = 0.5$ . Why this step? We obey the boxed rule — confirm $a > 1, b > 1$ before trusting the fraction.
Compare. $\hat{θ}_{MLE} = 8/10 = 0.8$ ; prior mean $0.2$ ; MAP $= 0.5$ . Why this step? To see the tug-of-war: MAP sits between data and prior.

Verify: $0.5$ lies strictly between $0.2$ and $0.8$ — a proper compromise. Sanity: the posterior is Beta $(10, 10)$ , symmetric, so its mode $= 0.5$ . ✓ This matches Figure 1's template: a right-leaning blue likelihood and a left-leaning pink prior blend to a centred yellow peak.

Worked example Example 2 — Cell B: watch the prior fade as data grows

Statement: The coin's true head-rate keeps producing $70%$ heads. Prior Beta $(3, 2)$ . Compute $\hat{θ}_{MAP}$ for $n = 10$ (so $k = 7$ ) and for $n = 1000$ (so $k = 700$ ).

Forecast: With little data the prior mean $0.6$ drags the estimate down; with mountains of data the estimate should snap to $0.7$ . Guess both numbers.

Small $n$ . Shape numbers $a = 7 + 3 = 10$ , $b = 3 + 2 = 5$ , both $> 1$ , so $\hat{θ}_{MAP} = \frac{10 - 1}{10 + 5 - 2} = \frac{9}{13} \approx 0.6923$ . Why this step? A first checkpoint on our formula; it reproduces the value from the parent note's coin example.
Large $n$ . $a = 700 + 3 = 703$ , $b = 300 + 2 = 302$ , both $> 1$ , so $\hat{θ}_{MAP} = \frac{703 - 1}{703 + 302 - 2} = \frac{702}{1003} \approx 0.6999$ . Why this step? Same formula; only $k, n$ change, so we watch the fixed prior constants ( $α, β$ ) become negligible.
Read the trend. $0.6923 \to 0.6999$ as $n$ multiplies by 100 — the estimate marches toward the MLE $0.7$ . Why this step? This is the general law: as $n \to \infty$ , MAP $\to$ MLE.

Verify: $702/1003 = 0.69990 \dots$ , and $∣0.6999 - 0.7∣ = 0.0001$ , far smaller than the small- $n$ gap $∣0.6923 - 0.7∣ = 0.0077$ . The prior's influence shrank ~77×. ✓

Figure 2 (this is Example 2). Each curve is the posterior for one of the three sample sizes in Example 2. As $n$ climbs (blue → pink → yellow), the peak slides from $\approx 0.69$ toward $0.70$ (the dashed white MLE line) and narrows — more data buys both accuracy and confidence. The dotted vertical lines are the three MAP peaks you computed above.

Worked example Example 3 — Cell C: zero successes (a degenerate input)

Statement: 0 heads in 5 flips ( $x$ is five tails). Prior Beta $(2, 2)$ . Find $\hat{θ}_{MAP}$ and note why MLE misbehaves.

Forecast: MLE screams $θ = 0/5 = 0$ ("this coin never lands heads") — an overconfident claim from 5 flips. Guess whether MAP is exactly $0$ or something positive.

MLE first. $\hat{θ}_{MLE} = 0/5 = 0$ . Why this step? To expose the pathology: zero probability of heads forever, from a handful of flips. This is overfitting to sparse data.
Check the edges, then MAP. Shape numbers $a = 0 + 2 = 2$ , $b = 5 + 2 = 7$ , both $> 1$ , so the interior formula applies: $\hat{θ}_{MAP} = \frac{2 - 1}{2 + 7 - 2} = \frac{1}{7} \approx 0.1429$ . Why this step? Beta $(2, 2)$ adds one "pseudo-head" and one "pseudo-tail," so $a > 1$ and the estimate stays off the boundary — no edge case here despite $k = 0$ .
Interpret. The prior refuses to believe "impossible heads," nudging the estimate to a modest $0.143$ . Why this step? This is exactly why we like priors: they regularize away absurd boundary answers.

Verify: $1/7 \approx 0.1429 > 0$ , strictly inside $(0, 1)$ , unlike MLE's boundary $0$ . Shape numbers $a = 2 > 1$ , $b = 7 > 1$ , so the interior formula was the right tool. ✓

Worked example Example 4 — Cell D: uniform prior must give back MLE

Statement: 7 heads in 10 flips. Prior Beta $(1, 1)$ (the uniform prior on $[0, 1]$ ). Show $\hat{θ}_{MAP} = \hat{θ}_{MLE}$ .

Forecast: Beta $(1, 1)$ is flat — every $θ$ equally believable beforehand. So MAP should reproduce MLE exactly: guess $0.7$ .

Recognize the flat prior. $θ^{1 - 1} (1 - θ)^{1 - 1} = θ^{0} (1 - θ)^{0} = 1$ for all $θ$ . Why this step? A constant prior adds a constant to the log-posterior, which cannot move the argmax.
Check the edges, then apply the formula. Shape numbers $a = 7 + 1 = 8$ , $b = 3 + 1 = 4$ , both $> 1$ , so $\hat{θ}_{MAP} = \frac{8 - 1}{8 + 4 - 2} = \frac{7}{10} = 0.7$ . Why this step? The " $+ α - 1$ " and " $+ β - 1$ " terms vanish for $α = β = 1$ , leaving exactly $k / n$ .
Confirm identity. $\hat{θ}_{MLE} = 7/10 = 0.7 = \hat{θ}_{MAP}$ . Why this step? This proves the parent claim: MLE is MAP with a uniform prior.

Verify: $7/10 = 0.7$ equals the MLE by construction. ✓ (See Maximum Likelihood Estimation (MLE).)

Worked example Example 5 — Cell E: mode ≠ mean, which do you report?

Statement: Posterior is Beta $(2, 5)$ . Report both the MAP (mode) and the posterior mean, and explain when the difference bites.

Forecast: For a skewed distribution these disagree. The mass leans toward small $θ$ , so mode $<$ mean or mode $>$ mean? Guess before computing.

Mode (MAP). Shape numbers $a = 2, b = 5$ , both $> 1$ , so mode $= \frac{a - 1}{a + b - 2} = \frac{1}{5} = 0.2$ . Why this step? MAP is the peak; the interior formula applies since $a > 1, b > 1$ .
Mean (MMSE point). Mean $= \frac{a}{a + b} = \frac{2}{7} \approx 0.2857$ . Why this step? The posterior mean is the estimate that minimizes mean-squared error — a different question than "where's the peak?"
Diagnose. $0.2 \neq = 0.2857$ : the long right tail drags the mean above the mode. Why this step? Reporting MAP when the audience expects a mean silently changes the statistic.

Verify: Mode $0.2$ , mean $0.2857$ ; the gap $0.0857 \neq = 0$ confirms asymmetry. Cross-check with the parent note's stated values ( $1/5$ and $2/7$ ). ✓

Gaussian / regularization cells

For continuous parameters with Gaussian noise, the likelihood peak is the sample mean $\overset{x}{ˉ}$ , and a Gaussian prior shrinks it toward the prior mean. This shrinkage is L2 regularization — see Regularization in ML, Ridge Regression.

Worked example Example 6 — Cell F: the L2 penalty is literally a prior

Statement: One observation $x = 2$ from $N (μ, 1)$ (so $x = {2}$ , $n = 1$ ). Gaussian prior $μ \sim N (0, σ_{0}^{2})$ with $σ_{0}^{2} = 4$ . Find $\overset{μ}{^}_{MAP}$ , and identify the ridge penalty $λ$ it corresponds to.

Forecast: MLE would say $\overset{μ}{^} = x = 2$ . The prior at $0$ shrinks it. Guess a number in $(0, 2)$ .

Log-posterior. $lo g P (μ ∣ x) = - \frac{1}{2} (x - μ)^{2} - \frac{μ ^{2}}{2 σ _{0}^{2}} + const$ . Why this step? Sum of log-likelihood (data fit) and log-prior (penalty on large $μ$ ).
Differentiate, set to 0. $(x - μ) - \frac{μ}{σ _{0}^{2}} = 0$ . Why this step? At the peak the slope is zero; this equation pins the optimal $μ$ .
Rearrange for $μ$ . Group the $μ$ terms: $x = μ + \frac{μ}{σ _{0}^{2}} = μ (1 + \frac{1}{σ _{0}^{2}}) = μ \cdot \frac{σ _{0}^{2} + 1}{σ _{0}^{2}}$ , so $\overset{μ}{^} = \frac{σ _{0}^{2} x}{σ _{0}^{2} + 1} = \frac{4 \cdot 2}{4 + 1} = \frac{8}{5} = 1.6$ . Why this step? We collect every $μ$ on one side and factor it out, so the equation becomes " $x = μ \times (constant)$ " and dividing gives $μ$ alone — pure algebra, no new idea.
Read off the penalty. The prior term $\frac{μ ^{2}}{2 σ _{0}^{2}} = λ μ^{2}$ with $λ = \frac{1}{2 σ _{0}^{2}} = \frac{1}{8} = 0.125$ . Why this step? This exposes MAP-with-Gaussian-prior as ridge regression with $λ = 1/ (2 σ_{0}^{2})$ .

Verify: $\overset{μ}{^} = 8/5 = 1.6 \in (0, 2)$ — shrunk from the MLE $2$ toward the prior mean $0$ , as promised. And $λ = 1/8 = 0.125$ . ✓

Worked example Example 7 — Cell G: real-world word problem

Statement: A factory sensor reads an object's length as $x = 5.2$ cm (measurement noise $σ = 1$ cm, so $N (μ, 1)$ ; here $x = {5.2}$ ). Blueprints say lengths cluster around $μ_{0} = 5.0$ cm with prior std $σ_{0} = 0.5$ cm ( $σ_{0}^{2} = 0.25$ ). What is the MAP length estimate?

Forecast: The reading $5.2$ vs. the blueprint $5.0$ : the prior is tight ( $σ_{0} = 0.5$ ), the measurement noise is looser ( $σ = 1$ ). So MAP should lean toward the blueprint. Guess closer to $5.0$ than to $5.2$ .

Translate English into a likelihood. "Sensor reads $5.2$ with noise $σ = 1$ " means the reading is drawn from $N (μ, 1)$ , so $lo g P (x ∣ μ) = - \frac{1}{2} (x - μ)^{2} + const$ . Why this step? The word "measurement noise" is code for a Gaussian likelihood centred on the true length $μ$ ; this is how we turn a sentence into math.
Translate English into a prior. "Blueprints cluster around $5.0$ with std $0.5$ " means $μ \sim N (μ_{0}, σ_{0}^{2})$ with $μ_{0} = 5.0, σ_{0}^{2} = 0.25$ , giving $lo g P (μ) = - \frac{( μ - μ _{0} ) ^{2}}{2 σ _{0}^{2}} + const$ . Why this step? Prior knowledge about the part's design becomes the prior distribution; a tight std means a confident prior.
Write the log-posterior. $lo g P (μ ∣ x) = - \frac{1}{2} (x - μ)^{2} - \frac{( μ - μ _{0} ) ^{2}}{2 σ _{0}^{2}} + const$ . Why this step? MAP = maximize log-likelihood + log-prior; we just add the two pieces from steps 1–2.
Differentiate and set to 0. $\frac{d}{d μ} = (x - μ) - \frac{μ - μ _{0}}{σ _{0}^{2}} = 0$ . Why this step? The peak of the posterior has zero slope; this equation locates it.
Rearrange for $μ$ . Multiply through by $σ_{0}^{2}$ : $σ_{0}^{2} (x - μ) - (μ - μ_{0}) = 0$ , i.e. $σ_{0}^{2} x + μ_{0} = μ (σ_{0}^{2} + 1)$ , so $\overset{μ}{^} = \frac{σ _{0}^{2} x + μ _{0}}{σ _{0}^{2} + 1}$ . Why this step? Clearing the fraction and collecting the $μ$ terms turns the equation into " $stuff = μ \times (constant)$ "; dividing isolates $μ$ . The result is a precision-weighted average of data and prior mean.
Plug numbers. $\overset{μ}{^} = \frac{0.25 \cdot 5.2 + 5.0}{0.25 + 1} = \frac{1.3 + 5.0}{1.25} = \frac{6.3}{1.25} = 5.04$ cm. Why this step? Direct substitution into the formula we just derived.

Verify: $6.3/1.25 = 5.04$ cm, and it lies in $[5.0, 5.2]$ , nearer the tighter-prior end. Precision weights: prior $1/0.25 = 4$ , data $1/1 = 1$ , ratio $4 : 1$ matches the strong pull to $5.0$ . Units: every term is in cm, so the answer is in cm. ✓

Worked example Example 8 — Cell H: prior with

$α < 1$ forces a boundary solution Statement: 3 heads in 3 flips. Prior Beta $(1, 1)$ (uniform). Also consider what a prior Beta $(0.5, 1)$ would do. Find $\hat{θ}_{MAP}$ in each. (Twist: the derivative may never equal zero inside $(0, 1)$ — where does the max live then?)

Forecast: Every flip was heads, so the likelihood $θ^{3}$ keeps rising toward $θ = 1$ . Guess: the peak is on the boundary, not an interior stationary point.

Uniform prior objective. Posterior $\propto θ^{3} (1 - θ)^{0} = θ^{3}$ for $θ \in [0, 1]$ ; shape numbers $a = 3 + 1 = 4$ , $b = 0 + 1 = 1$ . Why this step? With $k = n$ and a flat prior we get a pure increasing power of $θ$ , and $b = 1$ (not $> 1$ ) is a warning flag.
Apply the edge rule. Since $b = 1 \leq 1$ , the boxed rule says the peak sits on the right edge: $\hat{θ}_{MAP} = 1$ . (Check: $\frac{d}{d θ} θ^{3} = 3 θ^{2} = 0$ only at $θ = 0$ , a minimum, confirming no interior max.) Why this step? This is the whole point of Cell H — "set derivative to zero" fails, so we obey the boundary check instead.
The $α < 1$ variant. Prior Beta $(0.5, 1)$ gives posterior $\propto θ^{3 - 0.5} (1 - θ)^{0} = θ^{2.5}$ , shape numbers $a = 3.5 > 1$ , $b = 1 \leq 1$ — still a right-edge peak, $\hat{θ}_{MAP} = 1$ . Had we instead had $k = 0$ with an $α = 0.5$ prior, $a = 0.5 \leq 1$ would force $\hat{θ}_{MAP} = 0$ (left edge). Why this step? To show both edge directions: $a \leq 1 \Rightarrow 0$ , $b \leq 1 \Rightarrow 1$ .

Verify: $θ^{3}$ on $[0, 1]$ is largest at $θ = 1$ (value $1$ ), bigger than any interior point such as $0. 9^{3} = 0.729$ . So $\hat{θ}_{MAP} = 1$ . A stronger interior prior like Beta $(2, 2)$ would instead give $\frac{3 + 2 - 1}{3 + 2 + 2 - 2} = \frac{4}{5} = 0.8$ , softening the overconfident boundary answer. ✓

Recall Quick self-test (cover the answers)

What does MAP report, mode or mean? ::: The mode (peak) of the posterior. Beta $(1, 1)$ prior turns MAP into what? ::: Exactly the MLE (uniform prior). Before using the Beta mode formula, what must you check? ::: That both posterior shape numbers exceed 1 ( $a > 1$ and $b > 1$ ); otherwise the peak is on an edge ( $0$ or $1$ ). Zero heads in $n$ flips: why is MLE dangerous but MAP safe? ::: MLE gives $0$ (impossible-heads-forever); a Beta prior with $α > 1$ keeps $a > 1$ so MAP stays inside $(0, 1)$ . L2 regularization with strength $λ$ is MAP with which prior? ::: Gaussian $N (0, σ^{2})$ with $λ = 1/ (2 σ^{2})$ . When does the "set derivative to zero" step fail? ::: When $a \leq 1$ or $b \leq 1$ — the maximum is on the boundary, so check the endpoints.