What normalising constant c makes f(x) = c*x a valid density on [0, 2]?

## The two conditions that define a density A function f is a valid probability density function exactly when 1. `f(x) >= 0` for every x, and 2. the total area under f over the whole real line equals 1. Every normalisation problem is condition 2 solved for the unknown constant, with condition 1 used to pick the sign. ## Solving this one The density is `f(x) = c*x` on `[0, 2]` and 0 elsewhere, so only the interval `[0, 2]` contributes area. The area under a straight line through the origin from 0 to 2 is the area of a triangle with base 2 and height `2c`: ``` area = (1/2) * base * height = (1/2) * 2 * (2c) = 2c ``` Or by integration, the integral of `c*x` from 0 to 2 is `c * (2^2 / 2) = 2c`. Either way, ``` 2c = 1 -> c = 1/2 ``` Non-negativity settles the sign: on `[0, 2]` the variable x is non-negative, so `c*x >= 0` requires `c >= 0`, and `c = 1/2` is the answer rather than any negative root. The density is ``` f(x) = x/2 on [0, 2], 0 elsewhere ``` ## Sanity checks worth doing out loud - **Endpoint values.** `f(0) = 0` and `f(2) = 1`. The peak equals 1 here by coincidence of the numbers, not by rule — a density's peak is unconstrained. - **Shape.** A line rising from the origin: mass concentrates towards the right end of the support, so the median should sit above the midpoint 1. - **Area by geometry.** Triangle of base 2 and height 1 has area 1. Consistent. ## Getting the CDF Accumulate the area from the left. For `0 <= x <= 2`, the area under `t/2` from 0 to x is ``` F(x) = x^2 / 4 ``` and `F(x) = 0` for `x < 0`, `F(x) = 1` for `x > 2`. Check the boundaries: `F(0) = 0` and `F(2) = 4/4 = 1`. The CDF is non-decreasing on the support because `x^2` increases there — as it must be. With F in hand every probability question becomes arithmetic: - `P(X > 1.5) = 1 - F(1.5) = 1 - (2.25 / 4) = 1 - 0.5625 = 0.4375` - `P(0.5 < X <= 1.5) = F(1.5) - F(0.5) = 0.5625 - 0.0625 = 0.5` - The median solves `x^2 / 4 = 0.5`, giving `x = sqrt(2)` — about 1.414, above the midpoint, as the shape predicted. You can also get `P(X > 1.5)` directly as the area under `x/2` from 1.5 to 2, which is `(1/4) * (2^2 - 1.5^2) = (1/4) * 1.75 = 0.4375`. Two routes agreeing is a good final check. ## The mistakes this question is built to catch - **Normalising the peak instead of the area.** Setting `f(2) = 1` happens to give `c = 1/2` here, which is a trap: it is the right number for the wrong reason and collapses on any other support. On `[0, 4]` the area rule gives `c = 1/8`, while the peak rule would give `c = 1/4` and produce a function whose total area is 2. - **Forgetting the support.** Integrating `c*x` over the whole line diverges. The density is zero outside `[0, 2]`, and saying so is part of the answer. - **Summing instead of integrating.** Adding up f at a few x values is PMF thinking; a continuous density is integrated. - **Complementing wrongly.** `F(1.5) = 0.5625` is `P(X <= 1.5)`; the question asked for the other side. Reporting 0.5625 for `P(X > 1.5)` is the single most common slip. ## The general pattern For `f(x) = c * x^k` on `[0, b]` with `k > -1`, the area is `c * b^(k+1) / (k+1)`, so ``` c = (k + 1) / b^(k + 1) ``` Check it against this problem: `k = 1`, `b = 2` gives `c = 2 / 4 = 1/2`. Recognising this family means you can normalise any such density in one line rather than re-deriving the integral under pressure. ## How to say it in an interview "Area must be 1. The integral of `c*x` from 0 to 2 is `2c`, so `c = 1/2`, positive because the density cannot be negative. That makes `F(x) = x^2/4` on the support, so `P(X > 1.5) = 1 - 2.25/4 = 0.4375."