given a graph of density function, what can we conclude about expected value

Question

given the following graph (the density function), what can we conclude about the expected value? I got stuck a little bit with that question and I would appreciate your help!

I found out that C must be equal $\frac{2}{7}$ that the distribution function would be valid (=1).

furthermore, I know that $E[x]=\int_{-\infty}^{\infty}xf(x)dx$. but here it's equal to: $\int_{1}^{2}x(\frac{1}{7}x-\frac{1}{7})dx+\int_{2}^{4}x(\frac{2}{7})dx+\int_{4}^{6}x(-\frac{1}{7}x+\frac{4}{7})dx$ (after calculating the functions that describe each part of the graph).

Now - from evaluating the above sum of integrals I get the value of about 0.2 while it doesn't seem reasonable! from the graph it should be more then 3 as I understand.

would appreciate your help and advice!

Answer 1

Your density can be split in three parts $f_1$, $f_2$ and $f_3$:

$f_1(x) \cdot x = -2/7 \cdot x + 2/7 \cdot x^2$

$f_2(x) \cdot x = 2/7 \cdot x$

$f_3(x) \cdot x = 6/7\cdot x -1/7 \cdot x^2$

Then then antiderivatives are:

$I_1(x) = -2/14\cdot x^2 + 2/21 \cdot x^3$

$I_2(x) = 2/14 \cdot x^2$

$I_3(x) = 6/14 \cdot x^2 -1/21 \cdot x^3$

And finally the sum of the integrals:

$E = (-2/14 \cdot 2^2 + 2/21 \cdot 2^3 + 2/14 \cdot 1^2 - 2/21 \cdot 1^3) + ( 2/14 \cdot 4^2 - 2/14 \cdot 2^2) + (6/14 \cdot 6^2 -1/21 \cdot 6^3 - 6/14 \cdot 4^2 + 1/21 \cdot 4^3) = (-8+2+32-8+216-96)/14 + (16-2-216+64)/21 = 23/7 \approx 3.285714$

Answer 2

I have $c=\frac{2}{7}$ as well. The area is

$0.5c+2c+c=1$

$3.5c=1 \Rightarrow c=\frac{2}{7}$

The pdf is

$f(x)=\begin{cases} c\cdot (x-1), \ 1\leq x <2 \\ c , \quad \ 2\leq x <4 \\ c-1/2\cdot c\cdot (x-4), \ 4\leq x <6 \end{cases}$

The slope of the third interval is $\frac{f(x_1)-f(x_2)}{x_1-x_2}=\frac{c-0}{4-6}$

We can use the integral calculation as well to get the value of c.

$\int_1^2 c\cdot(x-1) \, dx+\int_2^4 c \, dx+\int_4^6 c-1/2\cdot c\cdot (x-4) \, dx=1$

$=\frac{c}{2}+2c+c=1$

And the expected value is

$E(X)=\int_1^2 2/7\cdot (x-1)\cdot x \, dx+\int_2^4 2/7\cdot x \, dx+\int_4^6 (2/7-1/2\cdot 2/7\cdot (x-4))\cdot x \, dx$

Thus $E(X)=\frac{23}{7}=3\frac27$