1

enter image description here

I know that this distribution is symmetrical about y axis since f(-x)=f(x) but how did we know that it was symmetrical about 0 an dthat it is a triangular distribution? Since I know that a triangular distribution has the following expression: enter image description here

And why did we infer from the symmetry that E(x)=μ=0 without computing any integral?

Rayri
  • 83
  • "symmetrical about $y$ axis" is the same as saying "symmetrical about $0$" since the $y$-axis is the line $x=0$; equivalently, saying $f(-x)=f(x)$ is the same as saying $f(0-x)=f(0+x)$, which demonstrates symmetry about $0$ – Henry Jul 24 '18 at 11:33

1 Answers1

1

when $x$ is positive, $x \in (0, \tau)$, the graph is a line with negative slope.

Now, flip it along the $y$-axis, you get an increasing line.

Together, you get a triangle.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149