Continuous distributions formulas

Use the total area equal to 1 to find an unknown constant in a given density.

F is non-decreasing and, for a continuous X, continuous. Its limits are 0 as x goes to -infinity and 1 as x goes to +infinity. The PDF is the derivative of the CDF wherever the derivative exists.

Valid for any continuous X. Also P(X > a) = 1 - F(a).

Solve F(x) = p for x. The p-th percentile (p as a fraction) is x_p; the median is the 50th percentile.

E[X] exists only when the integral of |x| f(x) converges. Some distributions have no mean, for example t with 1 degree of freedom. Convention used throughout: E[X^2] = integral of x^2 f(x) dx, and SD = sqrt(Var(X)).

Holds for any constants a and b. Adding a constant shifts the mean and leaves the variance unchanged. SD(aX + b) = |a| SD(X).

Requires P(X > a) > 0, and the conditional PDF is zero for x <= a. In general, conditioning on an event A with P(X in A) > 0 gives f(x)/P(X in A) for x in A. The result integrates to 1 over the conditioning region.

Defined only where f_Y(y) > 0. For a fixed y it is a density in x and integrates to 1. If X and Y are independent, f_{X|Y}(x | y) = f_X(x).

Zero density outside [a, b]; F is 0 below a and 1 above b. For a sub-interval inside [a, b], probability = length / (b - a).

The median equals the mean for the uniform distribution.

Support is x >= 0; lambda is the rate. P(X > x) is the survival function.

Mean and SD are equal. The median (about 0.693/lambda) is smaller than the mean 1/lambda, and the distribution is right-skewed. ln is the natural logarithm.

Holds for all s, t >= 0. The exponential is the only continuous distribution with this property. Time already waited does not change the distribution of the remaining wait.

Requires X1 and X2 independent. Reason: P(min > x) = P(X1 > x) P(X2 > x) = e^{-(lambda1 + lambda2) x}. The same holds for more than two: the rates add.

lambda > 0 is the average number of events in the interval considered. P(X >= 1) = 1 - e^{-lambda}.

Mean equals variance. If X1 ~ Poisson(l1) and X2 ~ Poisson(l2) are independent, then X1 + X2 ~ Poisson(l1 + l2).

Events occur independently at a constant average rate. The count in a window of length t has mean lambda*t, and the gaps between successive events are exponential with mean 1/lambda.

Use when n is large and p is small, so that np is moderate. The binomial's variance np(1-p) is close to np only when p is small.

Mean = median = mode = mu; variance sigma^2; symmetric about mu.

Divide by the standard deviation sigma, not the variance. P(X > a) = 1 - Phi((a - mu)/sigma).

Phi(0) = 0.5. Use this to turn negative z into positive z when reading a table.

Applies to every normal distribution, whatever mu and sigma are.

Requires X and Y independent and each normal, with a and b not both zero. Variances add with squared coefficients, even for a difference: Var(X - Y) = sigma_X^2 + sigma_Y^2.

Exact for a normal population. The standard deviation of X-bar (its standard error) is sigma/sqrt(n).

For X_1, ..., X_n i.i.d. from any population with mean mu and finite variance sigma^2. One consequence: Binomial(n, p) is approximately N(np, np(1-p)) when np and n(1-p) are both large. For integer counts, a continuity correction of 0.5 improves the approximation.

Takes only values >= 0. It is right-skewed and becomes more symmetric as k grows. The sum of independent chi-squared variables with k1 and k2 df is chi-squared with k1 + k2 df.

Follows from E[Z^2] = 1 and Var(Z^2) = 2 for one standard normal, together with the independence of the Z_i.

For a random sample of size n from N(mu, sigma^2). Here S^2 is the sample variance, which divides by n - 1. One df is lost because the sample mean is estimated from the data.

Symmetric about 0 and bell-shaped, with heavier tails than N(0,1). As k grows, t_k approaches N(0,1).

For k = 1 the mean does not exist; for k = 2 the variance is infinite. For k > 2 the variance exceeds 1, reflecting the heavier tails, and it tends to 1 as k grows.

For a random sample of size n from a normal population with unknown sigma, where S is the sample standard deviation (divisor n - 1). If sigma is known, use (X-bar - mu)/(sigma/sqrt(n)) ~ N(0,1) instead.