A probability mass function assigns a number to every outcome of a random variable. Almost every property of that variable, including total probability, expected value, and variance, is computed by adding those numbers up. Sigma notation is the shorthand every one of those sums gets written in, so reading it fluently now pays off immediately.
Sigma notation is a recipe for adding up a list of numbers in one compact line. It works like a loop: start at the first item, compute its value, and add that to a running total. Move to the next item and repeat, stopping once you reach the last one. The letter that counts which item you're on is only a placeholder, not an ingredient in the final dish. It tells you which term to compute next, then disappears once every term has been added. Changing where the counting starts or stops changes which terms are included, not what the instruction means.
\displaystyle\sum_{i=1}^{n} a_i means "add up a_i as i runs from 1 to n": start at i=1, evaluate a_i, add it to a running total, increase i by one, and repeat until i=n. The letter i is an index variable. It never appears in the final answer, only in the pattern of terms being added. Changing the starting or ending value changes which terms are included, not the notation's meaning. A finite sum is a sum with a fixed, known number of terms -- it starts at one index and stops at another, unlike an infinite series that never stops adding terms. Sigma notation is only shorthand for that addition; every finite sum can, in principle, be computed by brute force, one term at a time.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Discrete Mathematics I — Logic, Proof, and Structure
Compute \sum_{i=1}^{3} (2i+1).
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Why
(2(1)+1)+(2(2)+1)+(2(3)+1) = 3+5+7 = 15.
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From Discrete Mathematics I — Logic, Proof, and Structure
Compute \prod_{i=1}^{3} i.
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Why
1\times2\times3=6.
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Why
Partial sums building up
Toggle the partial-sums view and watch each new term stack on top of the running total.
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Which sum matches which sigma expression?
Read each written-out sum and work out the sigma expression it corresponds to, using the index, start, and stop values above.
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Recipe Evaluate a sum written in sigma notation Use when: you see \sum with a lower index, an upper index, and a formula for the general term, and need a single number Goal: the exact number the sum evaluates to
Quick recipe
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1
Identify the letter used as the index (commonly i, j, or k), the number written below \sum where counting starts, and the number written above where it stops.
Example
For \displaystyle\sum_{j=2}^{6}(2j-3): the index is j, it starts at 2, and it stops at 6.
Why: every later step needs the exact starting and stopping values to know how many terms to compute, and where to begin.
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2
Note the exact expression sitting to the right of \sum, since a single sign or exponent inside it changes every term it produces.
Example
The general term here is 2j-3: double the index and subtract three.
Why: this is the formula every specific term in the next step comes from, so it has to be copied exactly, sign included.
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3
Plug in the starting value, then each following integer, one at a time, up through the ending value, writing down each resulting number.
Example
j=2: 2(2)-3=1; j=3: 2(3)-3=3; j=4: 2(4)-3=5; j=5: 2(5)-3=7; j=6: 2(6)-3=9.
Why: the sum is defined as the total of exactly these terms, no more and no fewer.
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4
Add up the list of numbers from the previous step to get the final value.
Example
1+3+5+7+9=25
⚡ Shortcut
When the general term is linear in the index, each term is a fixed amount more (or less) than the one before it. Skip listing every term by hand: multiply the number of terms by the average of the first and last term. Here there are 5 terms, the first is 1 and the last is 9, so the sum is 5\cdot\dfrac{1+9}{2}=5\cdot5=25, matching the term-by-term total exactly. This only works when consecutive terms change by the same fixed amount every time; it isn't valid for a general term like j^2 or 2^j, where that difference isn't constant.
Why: adding term by term is exactly what the notation means, so it always gives the right value whether or not the shortcut also applies.
Common mistakes
- Treating the index variable as if it appears in the final answer, when it's only a placeholder that disappears once every term is added.
- Starting or stopping at the wrong index value, an off-by-one error, especially when the lower limit isn't 1.
- Mis-substituting into the general term, such as squaring or doubling the index inconsistently from one term to the next.
- Dropping a term at either end of the range, usually the last one.
Worked example
Evaluating a sum directly
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1
The same four steps as the recipe above, run on a new sum: compute \displaystyle\sum_{k=1}^{5} (2k+1).
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2
List each term: k=1{:}\ 3, k=2{:}\ 5, k=3{:}\ 7, k=4{:}\ 9, k=5{:}\ 11.
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3\boxed{3+5+7+9+11 = 35}
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Worked example
A sum of probabilities must total 1
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1
A random variable takes values 1, 2, 3, 4 with probabilities p_1, p_2, p_3, p_4. Write, in sigma notation, the one condition every valid probability assignment must satisfy.
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2
Every valid pmf's probabilities add to exactly 1, no more and no less.
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3\boxed{\sum_{i=1}^{4} p_i = 1}
Where this goes next
Chapter 4 (Discrete random variables) and Chapter 5 (Expectation)
\sum_i P(x_i) = 1 (validity of a pmf) and E[X] = \sum_i x_i P(x_i) (expected value) are both sigma-notation sums over a random variable's possible values. Neither formula is doing anything new mathematically; both are exactly this section's recipe, run once per outcome.
Check your understanding (counts toward your progress)
Evaluate the sum of squares \displaystyle\sum_{k=1}^{3} k^2.
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Why
1^2+2^2+3^2 = 1+4+9 = 14.
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Why
Square each value of k from 1 to 3 first, then add the three results.
Explain why \sum_{i=1}^{n} c (a constant c, added n times) equals nc, not just c.
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First mistake
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Why
Why
The index i doesn't appear inside the summand at all — every one of the n terms being added is the same constant c, so adding c to itself n times gives n times c, not a single c.
Look at whether the index i actually shows up anywhere inside the term being summed.
Which sum equals \displaystyle\sum_{i=1}^{5} i^2?
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This starts the squared term at 0² instead of 1², so it's missing the true first term and stops one term short of the true last term.
This squares 2 through 6, giving 4+9+16+25+36 — different terms entirely, not a re-index of 1 through 5.
This sums j itself, not j² — a different power entirely.
Why
Substituting i=j+1 (so j=i-1 runs from 0 to 4 as i runs from 1 to 5) turns i^2 into (j+1)^2: \sum_{j=0}^{4}(j+1)^2 = \sum_{i=1}^5 i^2.
Substitute i = j+1 so that j runs from 0 to 4 as i runs from 1 to 5.
In \displaystyle\sum_{i=1}^{n} p_i, what role does i play?
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The sum's VALUE is the final answer; i itself is just a counter used to build it.
i takes every value from 1 to n in turn during the sum — it isn't fixed at n.
The total (if this is a valid pmf) is the sum's value, 1 — not the index i itself.
Why
The index variable only names which term to compute next; it never appears in the evaluated total.
Practice: Reading and evaluating sigma notation (5 problems · doesn’t affect your progress)
Which expanded sum matches \displaystyle\sum_{i=1}^{4} 2^i?
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This lists i itself (1,2,3,4), not 2^i.
This stops at i=3, missing the last term 2^4=16.
This computes 2i (2 times i), not 2^i (2 to the power i) — a different operation entirely.
Why
2^i for i=1,2,3,4 gives 2^1+2^2+2^3+2^4 = 2+4+8+16 term by term.
2+4+8+16
Compute \displaystyle\sum_{i=1}^{6} i.
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Why
1+2+3+4+5+6 = 21.
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Why
Find the value of \displaystyle\sum_{k=2}^{5} k^2.
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Why
2^2+3^2+4^2+5^2 = 4+9+16+25 = 54.
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Why
The sum starts at k=2, not k=1 — don't include the k=1 term.
Evaluate \displaystyle\sum_{i=1}^{5} 4 (a constant, added once for every value of i).
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Why
The constant 4 is added 5 times (once per value of i from 1 to 5): 5×4=20.
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Why
The index i doesn't appear in the summand — every one of the 5 terms is just 4.
Evaluate \displaystyle\sum_{i=0}^{4} \binom{4}{i} using the fact that a full row of Pascal's triangle sums to a power of 2.
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Why
\sum_{i=0}^4\binom{4}{i} adds every entry of Pascal's triangle's row 4, which always totals 2^4=16.
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Why
This is exactly a full row of Pascal's triangle (row 4) added up.
Row n of Pascal's triangle always sums to 2^n.