Probability Density Function and Continuous Random Variable

The big shift from discrete to continuous. For a discrete variable you add probabilities: $P(X=1) + P(X=2) + \ldots$. For a continuous variable $X$ (heights, times, weights — anything measured on a scale) a single exact value has probability zero, so we work with a probability density function $f(x)$ and find probabilities as areas.

The two conditions for a valid pdf. A function $f(x)$ is a probability density function only if:

• Non-negative everywhere: $f(x) \ge 0$ for all $x$ (areas, hence probabilities, can never be negative).
• Total area one: $\int_{-\infty}^{\infty} f(x)\,dx = 1$ (some value must occur, so all the probability adds to $1$).

In practice $f(x) = 0$ outside a finite interval, so the integral only runs over where $f$ is non-zero.

Finding an unknown constant. Many questions give $f(x) = k\,(\ldots)$ and ask for $k$. You always use the total-area condition: set $\int_{\text{range}} f(x)\,dx = 1$ and solve.

Cambridge tip. Before doing anything else, sketch the density. A picture tells you the range, whether $f \ge 0$, and lets you sanity-check that areas land between $0$ and $1$.

Once you have a valid pdf, every probability is an integral: $P(a \le X \le b) = \int_a^b f(x)\,dx$

Key facts that simplify the work:

• $P(X = a) = \int_a^a f = 0$. So $P(X < a) = P(X \le a)$ — strict vs non-strict inequalities give the same answer.
• $P(X > a) = \int_a^{\text{top}} f(x)\,dx = 1 - P(X \le a)$, whichever is quicker.
• If $a$ or $b$ falls outside the range where $f \ne 0$, only integrate over the part of $[a,b]$ where $f$ is non-zero.

Piecewise densities — split the limits. Many 9231 questions define $f$ by different formulae on different intervals, for example $f(x) = \frac12 x$ on $[0,1]$ and $f(x) = \frac12(2-x)$ on $[1,2]$. To find a probability spanning the join, split the integral at the break-point and use the correct formula on each piece: $P(0 \le X \le 1.5) = \int_0^1 \tfrac12 x\,dx + \int_1^{1.5} \tfrac12(2-x)\,dx$

Never integrate one formula across a region where it does not apply — that is a guaranteed lost mark.

Cambridge tip. Write the limits explicitly and match each integrand to its interval. If you ever get a probability above $1$ or below $0$, stop — you have used a wrong limit or a wrong piece.

Expectation (the mean). The mean of a continuous variable weights each value $x$ by its density and integrates: $E(X) = \int x\, f(x)\,dx$ It is the "balance point" of the area — where the density would balance on a pivot.

Expectation of a function. For any function $g$, $E\big(g(X)\big) = \int g(x)\, f(x)\,dx$ The most important case is $g(x) = x^2$, giving $E(X^2) = \int x^2 f(x)\,dx$, which you need for the variance.

Variance (the spread). The working formula is: $\text{Var}(X) = E(X^2) - \big[E(X)\big]^2$ The order matters: find $E(X^2)$ and $E(X)$ separately, then subtract the square of the mean — not the mean itself. The standard deviation is $\sqrt{\text{Var}(X)}$.

A worked rhythm. For $f(x) = \frac{1}{9}x^2$ on $[0,3]$:

• $E(X) = \int_0^3 x\cdot\frac19 x^2\,dx = \frac19\big[\frac{x^4}{4}\big]_0^3 = \frac19\cdot\frac{81}{4} = \frac{9}{4}$.
• $E(X^2) = \int_0^3 x^2\cdot\frac19 x^2\,dx = \frac19\big[\frac{x^5}{5}\big]_0^3 = \frac19\cdot\frac{243}{5} = \frac{27}{5}$.
• $\text{Var}(X) = \frac{27}{5} - \big(\frac94\big)^2 = \frac{27}{5} - \frac{81}{16} = \frac{27}{80}$.

Cambridge tip. Write $\text{Var}(X) = E(X^2) - [E(X)]^2$ in full with the bracketed square BEFORE substituting. The single most common slip is subtracting $E(X)$ instead of $[E(X)]^2$.

These three "averages and positions" are each found differently.

Mode — the most likely value. The mode is where the density is highest. For a smooth density, solve: $f'(x) = 0 \quad\text{and check}\quad f''(x) < 0$ to confirm a maximum, and make sure it lies inside the range (if the peak is at an endpoint, compare endpoint heights instead). The mode is the $x$-value of the peak, not the height $f$ there.

Median — the halfway value. The median $m$ splits the total area into two equal halves: $\int_{-\infty}^{m} f(x)\,dx = 0.5$ Half the probability lies below $m$, half above. For a piecewise density, first work out the cumulative area at each break-point to decide which piece $m$ is in, then solve within that piece.

Percentiles. The $100p$-th percentile $x_p$ generalises the median: $\int_{-\infty}^{x_p} f(x)\,dx = p$ So the lower quartile uses $p = 0.25$, the median $p = 0.5$, the upper quartile $p = 0.75$.

Important. Mean, median and mode are generally different. They coincide only for symmetric densities. Do not assume the median equals the mean.

Cambridge tip. For a median in a piecewise density, always compute the area up to the join first. If that area is below $0.5$, the median is in the next piece — set "area-so-far $+$ second-piece integral $= 0.5$" and solve.