STAT 410 Cheat Sheet
Joint PDF RulesRR Variance & Covariance Bounds, Slicing & Above/Below
f (x, y) ≥ 0, f dx dy = 1, support = region If asked: “Find Var(X)”, Cov(X, Y ), or ρ Which var first? Integrate the one whose bounds
where f > 0 1. Compute E[X], E[Y ], E[X 2 ], depend on the other. If y-bounds involve x ⇒ in-
ZZ E[Y 2 ], E[XY ] tegrate dy first. Rectangle ⇒ either order works.
P (X, Y ) ∈ A = f (x, y) dx dy 2. Apply shortcut formulas: dy dx: fix x; y bottom→top; sweep x left→right.
A
Var(X) = E[X 2 ] − E[X]2 dx dy: fix y; x left→right; sweep y bottom→top.
If asked: “Find the value of C so that f (x, y) is a Cov(X, Y ) = E[XY ] − E[X]E[Y ] Above or below? Pick a test point inside the
valid joint pdf ” support; plug into the event inequality. If satisfied
RR Cov(X, Y ) ⇒ that side is your region.
1. Set f (x, y) dy dx = 1 ρ= p
2. Sketch support; use correct bounds Var(X) Var(Y ) Ex: X + Y < 3, test (0.5, 0.5): 1 < 3 ✓ ⇒ region
3. Integrate and solve for C
Key properties: is below y = 3 − x.
Common supports: Rectangle 0<x<a, 0<y<b; Var(aX + bY ) = a2 Var(X) + b2 Var(Y ) + Piecewise: if target value crosses breakpoint, split
Triangle 0<y<x<1; Slanted 0<x<a, 0<y<c+x. 2ab Cov(X, Y ) the integral there.
Marginal Densities Var(aX + b) = a2 Var(X) (constants shift mean, Region Probability Method
If asked: “Find the marginal pdf of X, fX (x)” not variance) If asked: “Find P (X + Y < c)”, “Find P (Y >
1. Sketch support; fix x Cov(X, X) = Var(X); Cov(X, Y ) = Cov(Y, X) kX)”, “Find P (XY < c)”
2. Identify y-bounds that R depend on x Cov(aX + b, cY + d) = ac Cov(X, Y ) 1. Rewrite event as inequality; identify boundary
3. Integrate: fX (x) = (y range|x) f (x, y) dy curve
Conditional Expectation
4. State support of X 2. Sketch support; draw boundary on it
If asked: “Find E(Y | X = x)” or “Find E(X |
Same process for fY (y) but fix y, integrate over x. 3. Find intersections of boundary with support
Y = y)”
edges
If support is slanted (e.g. 0 < y < 2 + x), fY (y) is Translating a sentence → notation:
4. Set integration limits (split if top/bottom
piecewise — find where x-bounds change. Scan for the word “given” — everything after it
changes)
Piecewise example: support 0<x<2, 0<y<2+x. goes after the bar (it’s fixed). What you’re find-
5. Integrate f over the region
For fY (y): fix y, find x-range. If 0<y<2: x goes ing the expected value of goes before the bar.
0 to 2. If 2<y<4: x goes y−2 to 2. Breakpoint at Boundary types: x + y = c: line; y = kx: line thru
“Expected grade, given hours = s” ⇒ E(Grade |
y=2. origin; xy = c: hyperbola.
Hours = s)
Upper bound = min(event bdy, support top).
Conditional Density “E(Y | X = x) written directly” ⇒ Y before bar
Split where equal.
If asked: “Find P (Y > a | X = x0 )” or “Find (integrate), X = x after bar (fixed).
P (X < a | Y = y0 )” Steps: How to Choose Bounds from Sketch
1. Pick a fixed value of the outer variable
1. Find fX (x0 ) using marginal (or given) 1. Find fY |X (y|x) = f (x, y)/fX (x) 2. Draw a vertical (or horizontal) slice through the
2. Form fY |X (y | x0 ) = f (x0 , y) / fX (x0 ) 2. Plug fixedRvalue into density AND y-bounds support
3. Plug in the given value; adjust bounds to con- 3. Compute y · fY |X (y|x) dy; answer is in x only 3. Read where slice enters (lower bound) and exits
ditional support If fY piecewise ⇒ two-piece answer, one per range (upper bound)
4. Integrate over the event region of y. RR 4. Sweep outer variable across the full support
f (x, y) f (x, y) x f dx dy 5. Split if the upper bound changes form at some
fY |X (y|x) = , fX|Y (x|y) = Region: E(X | A) = A
fX (x) fY (y) P (A) point
If fY is piecewise: use the piece matching y0 . Law of Total Variance: Var(Y ) = E[Var(Y |
Ex: support 0 < x < 2, 0 < y < 2 + x. Fix x = 1:
Shortcut: if fY |X looks like a known pdf (Exp, X)] + Var(E[Y | X])
slice enters at y = 0, exits at y = 3. As x sweeps
Uniform), read off mean directly. Ex: fY |X = “Variance of the conditional mean + mean of the 0 → 2, top always y = 2 + x — no split needed.
λe−λy ⇒ E(Y |X = x) = 1/λ. conditional variance.”
Independence Test FW (w) — CDF of W = X + Y
Expectation If asked: “Are X and Y independent? Justify your If asked: “Use the CDF approach to find FW (w)”
If asked: “Find E(X)” or “Find E(g(X, Y ))” answer.” / “Find P (X + Y ≤ w)”
1. Find marginal if needed 1. Check support shape — if non-rectangular 1. Sketch support; draw line y = w − x (keep w
2. Multiply variable by its pdf (slanted/triangular): NOT independent, symbolic)
3. Integrate over support stop 2. Find breakpoint: plug corners into x + y to get
Z Z
2. If rectangular: check f (x, y) = fX (x) · fY (y) case boundaries
E[X] = x fX (x) dx, E[Y ] = y fY (y) dy
3. State conclusion with justification 3. For each case integrate f over region below line
ZZ 4. Differentiate FW′ (w) = f (w)
W
E[g(X, Y )] = g(x, y) f (x, y) dx dy Cov(X, Y ) ̸= 0 ⇒ not independent (but = 0 alone
isn’t enough). Fast breakpoint: list support corners; their x + y
R
If independent: E(XY ) = E(X)E(Y ); Cov = 0; sums = split values.
LOTUS: E[g(X)] = g(x)fX (x) dx — no need to
find pdf of g(X). ρ = 0; Var(X + Y ) = Var(X) + Var(Y ). fW (w) — PDF of W = X + Y (convolution)
Linearity: E[aX + bY + c] = aE[X] + bE[Y ] + c Factorization test: if f (x, y) = g(x)h(y) for all If asked: “Use the convolution approach to find
— always holds, even if dependent. (x, y) in a rectangular support, then independent. fW (w)”
R
The functions g, h don’t need to be proper pdfs — 1. Write fW (w) = f (x, w − x) dx
Law of Total Expectation: E[Y ] = E E(Y |
R
X) = E(Y | X=x) fX (x) dx. just need the factorization. 2. Sub y = w−x into all support inequalities; solve
Value vs. Region Conditioning each for x
Value (X = c): use conditional pdf ⇒ single inte- 3. Take the overlap of all resulting x-intervals as
gral. your bounds
Region (Y < b): P (A | B) = P (A ∩ B)/P (B) ⇒ 4. Integrate; answer is a function of w (will be
double integrals. piecewise)
= c ⇒ cond. pdf. <, >, ≤, ≥⇒ region prob. Key: convolution ̸= CDF method — same final
Variable before the bar = variable you integrate. answer, different route. R
Conditioning changes bounds! Plug x0 into If independent: fW (w) = fX (x) fY (w − x) dx
both the density and the support inequalities. (product of marginals).
Ex: support 0 < y < 2 + x, given X = 1: y-bounds Support of W : w ∈ [min(x+y), max(x+y)] from
become 0 < y < 3. support corners.
Joint PDF RulesRR Variance & Covariance Bounds, Slicing & Above/Below
f (x, y) ≥ 0, f dx dy = 1, support = region If asked: “Find Var(X)”, Cov(X, Y ), or ρ Which var first? Integrate the one whose bounds
where f > 0 1. Compute E[X], E[Y ], E[X 2 ], depend on the other. If y-bounds involve x ⇒ in-
ZZ E[Y 2 ], E[XY ] tegrate dy first. Rectangle ⇒ either order works.
P (X, Y ) ∈ A = f (x, y) dx dy 2. Apply shortcut formulas: dy dx: fix x; y bottom→top; sweep x left→right.
A
Var(X) = E[X 2 ] − E[X]2 dx dy: fix y; x left→right; sweep y bottom→top.
If asked: “Find the value of C so that f (x, y) is a Cov(X, Y ) = E[XY ] − E[X]E[Y ] Above or below? Pick a test point inside the
valid joint pdf ” support; plug into the event inequality. If satisfied
RR Cov(X, Y ) ⇒ that side is your region.
1. Set f (x, y) dy dx = 1 ρ= p
2. Sketch support; use correct bounds Var(X) Var(Y ) Ex: X + Y < 3, test (0.5, 0.5): 1 < 3 ✓ ⇒ region
3. Integrate and solve for C
Key properties: is below y = 3 − x.
Common supports: Rectangle 0<x<a, 0<y<b; Var(aX + bY ) = a2 Var(X) + b2 Var(Y ) + Piecewise: if target value crosses breakpoint, split
Triangle 0<y<x<1; Slanted 0<x<a, 0<y<c+x. 2ab Cov(X, Y ) the integral there.
Marginal Densities Var(aX + b) = a2 Var(X) (constants shift mean, Region Probability Method
If asked: “Find the marginal pdf of X, fX (x)” not variance) If asked: “Find P (X + Y < c)”, “Find P (Y >
1. Sketch support; fix x Cov(X, X) = Var(X); Cov(X, Y ) = Cov(Y, X) kX)”, “Find P (XY < c)”
2. Identify y-bounds that R depend on x Cov(aX + b, cY + d) = ac Cov(X, Y ) 1. Rewrite event as inequality; identify boundary
3. Integrate: fX (x) = (y range|x) f (x, y) dy curve
Conditional Expectation
4. State support of X 2. Sketch support; draw boundary on it
If asked: “Find E(Y | X = x)” or “Find E(X |
Same process for fY (y) but fix y, integrate over x. 3. Find intersections of boundary with support
Y = y)”
edges
If support is slanted (e.g. 0 < y < 2 + x), fY (y) is Translating a sentence → notation:
4. Set integration limits (split if top/bottom
piecewise — find where x-bounds change. Scan for the word “given” — everything after it
changes)
Piecewise example: support 0<x<2, 0<y<2+x. goes after the bar (it’s fixed). What you’re find-
5. Integrate f over the region
For fY (y): fix y, find x-range. If 0<y<2: x goes ing the expected value of goes before the bar.
0 to 2. If 2<y<4: x goes y−2 to 2. Breakpoint at Boundary types: x + y = c: line; y = kx: line thru
“Expected grade, given hours = s” ⇒ E(Grade |
y=2. origin; xy = c: hyperbola.
Hours = s)
Upper bound = min(event bdy, support top).
Conditional Density “E(Y | X = x) written directly” ⇒ Y before bar
Split where equal.
If asked: “Find P (Y > a | X = x0 )” or “Find (integrate), X = x after bar (fixed).
P (X < a | Y = y0 )” Steps: How to Choose Bounds from Sketch
1. Pick a fixed value of the outer variable
1. Find fX (x0 ) using marginal (or given) 1. Find fY |X (y|x) = f (x, y)/fX (x) 2. Draw a vertical (or horizontal) slice through the
2. Form fY |X (y | x0 ) = f (x0 , y) / fX (x0 ) 2. Plug fixedRvalue into density AND y-bounds support
3. Plug in the given value; adjust bounds to con- 3. Compute y · fY |X (y|x) dy; answer is in x only 3. Read where slice enters (lower bound) and exits
ditional support If fY piecewise ⇒ two-piece answer, one per range (upper bound)
4. Integrate over the event region of y. RR 4. Sweep outer variable across the full support
f (x, y) f (x, y) x f dx dy 5. Split if the upper bound changes form at some
fY |X (y|x) = , fX|Y (x|y) = Region: E(X | A) = A
fX (x) fY (y) P (A) point
If fY is piecewise: use the piece matching y0 . Law of Total Variance: Var(Y ) = E[Var(Y |
Ex: support 0 < x < 2, 0 < y < 2 + x. Fix x = 1:
Shortcut: if fY |X looks like a known pdf (Exp, X)] + Var(E[Y | X])
slice enters at y = 0, exits at y = 3. As x sweeps
Uniform), read off mean directly. Ex: fY |X = “Variance of the conditional mean + mean of the 0 → 2, top always y = 2 + x — no split needed.
λe−λy ⇒ E(Y |X = x) = 1/λ. conditional variance.”
Independence Test FW (w) — CDF of W = X + Y
Expectation If asked: “Are X and Y independent? Justify your If asked: “Use the CDF approach to find FW (w)”
If asked: “Find E(X)” or “Find E(g(X, Y ))” answer.” / “Find P (X + Y ≤ w)”
1. Find marginal if needed 1. Check support shape — if non-rectangular 1. Sketch support; draw line y = w − x (keep w
2. Multiply variable by its pdf (slanted/triangular): NOT independent, symbolic)
3. Integrate over support stop 2. Find breakpoint: plug corners into x + y to get
Z Z
2. If rectangular: check f (x, y) = fX (x) · fY (y) case boundaries
E[X] = x fX (x) dx, E[Y ] = y fY (y) dy
3. State conclusion with justification 3. For each case integrate f over region below line
ZZ 4. Differentiate FW′ (w) = f (w)
W
E[g(X, Y )] = g(x, y) f (x, y) dx dy Cov(X, Y ) ̸= 0 ⇒ not independent (but = 0 alone
isn’t enough). Fast breakpoint: list support corners; their x + y
R
If independent: E(XY ) = E(X)E(Y ); Cov = 0; sums = split values.
LOTUS: E[g(X)] = g(x)fX (x) dx — no need to
find pdf of g(X). ρ = 0; Var(X + Y ) = Var(X) + Var(Y ). fW (w) — PDF of W = X + Y (convolution)
Linearity: E[aX + bY + c] = aE[X] + bE[Y ] + c Factorization test: if f (x, y) = g(x)h(y) for all If asked: “Use the convolution approach to find
— always holds, even if dependent. (x, y) in a rectangular support, then independent. fW (w)”
R
The functions g, h don’t need to be proper pdfs — 1. Write fW (w) = f (x, w − x) dx
Law of Total Expectation: E[Y ] = E E(Y |
R
X) = E(Y | X=x) fX (x) dx. just need the factorization. 2. Sub y = w−x into all support inequalities; solve
Value vs. Region Conditioning each for x
Value (X = c): use conditional pdf ⇒ single inte- 3. Take the overlap of all resulting x-intervals as
gral. your bounds
Region (Y < b): P (A | B) = P (A ∩ B)/P (B) ⇒ 4. Integrate; answer is a function of w (will be
double integrals. piecewise)
= c ⇒ cond. pdf. <, >, ≤, ≥⇒ region prob. Key: convolution ̸= CDF method — same final
Variable before the bar = variable you integrate. answer, different route. R
Conditioning changes bounds! Plug x0 into If independent: fW (w) = fX (x) fY (w − x) dx
both the density and the support inequalities. (product of marginals).
Ex: support 0 < y < 2 + x, given X = 1: y-bounds Support of W : w ∈ [min(x+y), max(x+y)] from
become 0 < y < 3. support corners.
