Continuous random variables flashcards

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Continuous random variable
A random variable that can take any value in an interval (a time, length, mass), described by a probability density function.
Probability density function (pdf) f(x)
A function with f(x) ≥ 0 for all x and total area 1, such that P(a < X < b) = ∫ₐᵇ f(x) dx.
Cumulative distribution function (cdf) F(x)
F(x) = P(X ≤ x) = ∫ f(t) dt from −∞ to x: the area under the pdf to the left of x.
Median of a continuous random variable
The value m with F(m) = 0.5: half the area under the pdf lies on each side.
Mode of a continuous random variable
The value of x at which f(x) is greatest (a turning point or an end point of the range).
Quartiles and percentiles
Q₁: F(Q₁) = 0.25; Q₃: F(Q₃) = 0.75; the nth percentile p: F(p) = n/100.
P(a < X < b) for a continuous random variable
∫ₐᵇ f(x) dx = F(b) − F(a). (No — learn it)
F(x) in terms of f
F(x) = ∫ f(t) dt from −∞ to x. (Yes)
f(x) in terms of F
f(x) = F′(x) = dF/dx. (No — derive it)
E(X) for a continuous random variable
E(X) = ∫ x f(x) dx over the whole range. (Yes)
Var(X) for a continuous random variable
Var(X) = ∫ x² f(x) dx − μ² = E(X²) − [E(X)]². (Yes)
E(g(X)) for a continuous random variable
E(g(X)) = ∫ g(x) f(x) dx. (Yes)
E(aX + b) and Var(aX + b)
E(aX + b) = aE(X) + b; Var(aX + b) = a²Var(X). (No — learn it)
Shape of a cdf
Starts at 0, never decreases, has no jumps, and reaches 1 at the top of the range.
Probability on a sketch of a pdf
The area under the curve between the two values.
pdf with a long tail to the right
Positive skew: typically mode < median < mean.
pdf with a long tail to the left
Negative skew: typically mean < median < mode.
Symmetric single-peaked pdf
Mean = median = mode, at the line of symmetry.
pdf that increases over its whole range
The mode is the upper end point of the range.
Why is P(X = a) = 0 for continuous X?
It is the area of a strip of zero width: ∫ₐᵃ f(x) dx = 0. So < and ≤ give the same probability.
Can f(x) be greater than 1?
Yes. f is a density, not a probability; only the total area must be 1 (e.g. f = 2.5 on [0, 0.4]).
Building F for a piecewise pdf
On each piece: F(x) = (area of all earlier pieces) + ∫ from the join to x of f(t) dt.
Conditions for finding constants in a cdf
F = 0 at the lowest value, F = 1 at the highest, and equal values on both sides of each join.
Finding a median from a piecewise cdf
Compare 0.5 with F at each join to choose the piece, then solve F(m) = 0.5 there; reject roots outside it.
Justifying skewness
Quote the mean, median and mode and compare their order; the comparison is the reason.
f(x) = kx² for 0 ≤ x ≤ 3, 0 otherwise. Find k.
∫₀³ kx² dx = 9k = 1, so k = 1/9.
f(x) = x/8 for 0 ≤ x ≤ 4. Find P(X > 2).
∫₂⁴ x/8 dx = (16 − 4)/16 = 3/4.
Find F(x) for f(x) = x/8, 0 ≤ x ≤ 4.
F(x) = 0 for x < 0; x²/16 for 0 ≤ x ≤ 4; 1 for x > 4.
F(x) = (x³ − 8)/56 for 2 ≤ x ≤ 4. Find f(x).
f(x) = 3x²/56 for 2 ≤ x ≤ 4, and 0 otherwise.
f(x) = 3x²/8 for 0 ≤ x ≤ 2. Find E(X) and Var(X).
E(X) = ∫ 3x³/8 dx = 1.5; E(X²) = 2.4; Var(X) = 2.4 − 2.25 = 0.15.
E(X) = 4, Var(X) = 2. Find E(3X + 1) and Var(3X + 1).
E = 3 × 4 + 1 = 13; Var = 3² × 2 = 18.
F(x) = x²/16 for 0 ≤ x ≤ 4. Find the median.
m²/16 = 0.5 ⇒ m = √8 = 2√2 ≈ 2.83.
f(x) = ¾x(2 − x) for 0 ≤ x ≤ 2. Find the mode, mean and median.
Symmetric about x = 1 (f′ = 0 at x = 1), so mode = mean = median = 1.
A pdf or cdf written for one interval only
Incomplete: add '0 otherwise' to f, and the 0 and 1 lines to F.
Second piece of F missing the earlier area
F would not reach 1 at the top. Add F at the join before integrating the next piece.
Var(X) given as ∫x²f(x) dx
That is E(X²). Subtract [E(X)]².
Mode found only by f′(x) = 0
Check the end points: an increasing or decreasing pdf has its mode at an end of the range.
Rounding E(X) before finding Var(X)
Keep E(X) and E(X²) exact; rounding the mean can change the variance.

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