Assemble a continuous expectation using partial fractions

Join the existing foundations

Density normalisation and expected value are shared probability foundations; partial fractions and rational integration are already taught. Here the learner must assemble them. No independence assumption is needed for the mean of a single variable

Check density conditions first

A density must be nonnegative and integrate to 1 over its support. For the density below, the pole at −3 is outside the interval and the denominator is positive. Normalisation will give a positive K; the density is zero outside the stated support

Work the normalising integral

Integrate over the support, then set the total probability to 1

Give the normalising constant K for this density, whose total integral is 1.

Choose the expectation integral

The mean integrates each value weighted by its density over the complete support. Integrating the density alone gives total probability 1; integrating the values alone does not weight them by density

Which integrand belongs inside the expectation integral?

Decompose the assembled integrand

Rewrite the numerator, then divide each term by the squared denominator. Multiplying both sides by the squared denominator verifies the identity on the support. This decomposition follows from the expectation integrand, not from the density alone

Give B in the identity below

Evaluate both endpoints exactly

Integrate the decomposed expression, substitute both endpoints, then multiply by the normalising constant. The logarithm argument is positive throughout the support, so no absolute-value branch is crossed

For the density below, write the mean in the stated form, where R is the upper endpoint value of x+3 divided by its lower endpoint value. Give P.

For the density below, give Q in the stated form, where R is the upper endpoint value of x+3 divided by its lower endpoint value

Give R in the form below. R is the value of x+3 at the upper support endpoint divided by its value at the lower endpoint; the support is from −1 to 2.

Construct and justify the whole expectation, ungraded

For the density below, establish its validity, form the expectation integral, show the partial-fraction identity and evaluate both endpoints. Explain why a mean can be negative for a different shifted support even though density is nonnegative, and why no independence assumption is involved

Model: distinguish probability normalisation from the first moment

The density is K times the reciprocal of the square of x+3 from −1 to 2, and zero elsewhere. The pole at −3 lies outside the support, so a positive K gives a nonnegative density. Normalising gives the constant below. Form the first moment, decompose its integrand and evaluate at both endpoints as shown. The mean is approximately 0.0543, within the support. An expectation is a weighted average of signed values, so negative support values can yield a negative mean; that does not make the density negative. This calculation concerns one variable and makes no statement about independence of separate variables