A continuous random variable X has density f(x) = x for 0 < x <= 1, f(x) = 2 - x for 1 <= x < 2, and 0 otherwise. Find E(X) and Var(X).

Step-by-step solution

1. 1

   Check that $f$ really is a density. A legitimate pdf is non-negative and integrates to $1$:

   $\int_0^1x\,dx+\int_1^2(2-x)\,dx=\frac12+\left[2x-\frac{x^2}{2}\right]_1^2=\frac12+\left(2-\frac32\right)=\frac12+\frac12=1$

   Geometrically this is a triangle with base $2$ and height $1$, area $\frac12\cdot2\cdot1=1$. Confirming this first is cheap insurance against a mis-copied piece.

2. 2

   Compute $E(X)$ piece by piece. Because the density is defined in two pieces, the expectation integral splits at $x=1$:

   $E(X)=\int_0^1x\cdot x\,dx+\int_1^2x(2-x)\,dx=\int_0^1x^{2}dx+\int_1^2\left(2x-x^{2}\right)dx$

   $=\frac13+\left[x^{2}-\frac{x^{3}}{3}\right]_1^2=\frac13+\left(\frac43-\frac23\right)=\frac13+\frac23=1$

3. 3

   Notice the shortcut you could have used. The triangle is symmetric about $x=1$: $f(1-u)=1-u$ and $f(1+u)=1-u$ for $0\le u\le1$. For any density symmetric about a point, the mean is that point, so $E(X)=1$ without integrating at all. The integral above is then just a check.

4. 4

   Compute $E(X^{2})$, which symmetry cannot give you.

   $E(X^{2})=\int_0^1x^{3}dx+\int_1^2x^{2}(2-x)\,dx=\frac14+\left[\frac{2x^{3}}{3}-\frac{x^{4}}{4}\right]_1^2$

   Evaluating the bracket at $2$ gives $\frac{16}{3}-4=\frac43$; at $1$ it gives $\frac23-\frac14=\frac{5}{12}$. So the second integral is $\frac43-\frac{5}{12}=\frac{11}{12}$ and

   $E(X^{2})=\frac14+\frac{11}{12}=\frac{3}{12}+\frac{11}{12}=\frac{14}{12}=\frac{7}{6}$

5. 5

   Apply the shortcut formula for the variance.

   $D(X)=E(X^{2})-\left[E(X)\right]^{2}=\frac76-1^{2}=\frac{7}{6}-\frac{6}{6}=\frac16\approx0.1667$

   The standard deviation is $\sqrt{1/6}=0.408$.

6. 6

   Check that the answer is plausible. All the probability lives in $(0,2)$, an interval of length $2$, so the standard deviation must be well under $1$; $0.408$ is comfortable. It is also smaller than the $0.577$ of a uniform distribution on $(0,2)$, which is exactly right — the triangular density concentrates mass near its centre instead of spreading it evenly.