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Find the expectation of $X$ given that:

$F_X(x) = 0$ if $x < 0$;

$F_X(x) = 2x-x^2$ if $0\le x \le 1$;

$F_X(x) = 1$ if $x \ge 1$

I don't know what to do. Can you help me?

user13
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1 Answers1

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Two possibilities:

  • check this is a continuous distribution, differentiate to find the density $f(x)=F'(x)$ and then find $\int\limits_{-\infty}^\infty x\, f(x)\, dx$

  • check this is a non-negative distribution, find the survival function $S(x)=1-F(x)$ and then find $\int\limits_0^\infty S(x)\, dx$

Henry
  • 157,058
  • $P(x<0) = 0$ so I guess, it is non-negative. I think, it is continuous in $[0,1]$ but not in $x\ge1$ or $x<0$. The $1-F(x)$ will be for $x<0$: $1-0 = 1$, for $x \in [0,1]$ $1-2x+x^2$ and for $x \ ge 1$: $1-(1-F(1)) = 1$. Is this right so far? – user13 Sep 24 '19 at 22:13
  • No - it is not correct. You don't care about $S(x)$ for $x \lt 0$, and more importantly $S(x)=0$ for $x \gt 1$ – Henry Sep 25 '19 at 00:00
  • Can you explain why I don't care for $x<0$? – user13 Sep 25 '19 at 00:03
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    Because you checked this is a non-negative distribution, and the integral is over the positive reals. If you had taken the more general expression for the expectation of $\int\limits_0^\infty S(x), dx - \int\limits_{-\infty}^0 F(x), dx$ then you might care but here $F(x)=0$ for $x \lt 0$ and the second part disappears – Henry Sep 25 '19 at 00:09