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The cdf of a random variable $X$ is given by the following.

\[ F_X(x) = \begin{cases} 0 & x < 0\\\\ \frac x2 & x \in [0,1)\\\\ \frac 23 & x \in [1,2)\\\\ \frac{11}{12} & x \in [2,3) \\\\ 1 & x \ge 3 \end{cases} \] How do I calculate the expectation of the random variable $X$?

3 Answers 3

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$$\int_0^\infty (1 - F(x))\ dx$$

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    Would you elaborate? How should I integrate since F(x) can take many different forms depending on x.2012-09-24
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    @idealistikz just break it down into 3 parts, depending on the interval, and the integral over the whoel thing is the sum of the intergals over individual parts.2012-09-24
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    Why do I break it into 3 parts when there are 5 conditions? Do you mean 5 parts?2012-09-24
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    @idealistikz: Because the integral over the last part is clearly $0$, and the first part is irrelevant: the probability there is $0$.2012-09-24
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    Why is the integral over the last part 0? When x>=3, isn't the integral for that part x?2012-09-24
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    @idealistikz: $1-F_X(x)=1-1=0$ for $x\ge 3$.2012-09-24
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    This works only because the random variable is everywhere nonnegative. Maybe there's something to be said for an answer that works more generally.2012-09-24
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    @MichaelHardy: true. More generally, $\int_0^\infty (1-F(x))\ dx - \int_{-\infty}^0 F(x)\ dx$2012-09-24
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    The formula used here for the expectation of a nonnegative random variable has come up on this site many times for a variety of different problems. Someone must like to pick problems that exploit that formula for homework.2012-09-24
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    @BrianM.Scott, since the limit begins at 3, shouldn't I consider it because it doesn't cancel out?2012-09-25
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    @idealistikz: What doesn’t cancel out? $\int_3^\infty(1-1)dx=0$.2012-09-25
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    I was under the impression I calculate the integral of each interval separately, add them together, and subtract the total from the integral of 1 from 0 to infinity. Rather, am I supposed calculate the integral of 1-F(x) at each interval and sum the total?2012-09-25
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There is another, definitely less good way, to find the expectation. Let us analyze the distribution of the random variable $X$. The part before $0$ is harmless.

For $0\le x\lt 1$, the cumulative distribution function is $x/2$, nice and familiar, the density function is the derivative of the the cdf, which is $1/2$.

At $1$, and all the way up to but not including $2$, the cdf is $2/3$. So there is a sudden jump at $x=1$. As we approach $1$ from the left, the cdf approaches $1/2$, but all of a sudden it is $2/3$ at $1$, and then stays at that all the way to, but not including $2$. What that means is that there is a discrete "weight" of $2/3-1/2$ at $x=1$: $\Pr(X=1)=2/3-1/2=1/6$.

The cdf takes another sudden jump to $11/12$ at $2$. That means we have a weight of $11/12-2/3$, that is, $3/12$, at $x=2$.

Finally, there is another weight of $1/12$ at $x=3$.

To sum up, this is a mixed continuous-discrete situation: there is a continuous uniform distribution, density $1/2$, between $0$ and $1$. In addition, $\Pr(X=1)=1/6$, $\Pr(X=2)=3/12$, and $\Pr(X=3)=1/12$. The moment about the origin (mean) is therefore $$\int_0^1 x\cdot\frac{1}{2}\,dx+ 1\cdot\frac{1}{6}+2\cdot \frac{3}{12}+3\cdot \frac{1}{12}.$$

I think this simplifies to $\dfrac{7}{6}$. You might want to compare that with the result you get from doing it the full integration way. The latter approach is the one you should become comfortable with.

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You can come up with a piecewise defined density function by differentiating, but that fails to account for the jump discontinuities at $x=1$, $x=2$, and $x=3$. You get $$ f(x) = 1/2\text{ if }1$X$, we then have $$ \begin{cases} \Pr(X\in A) = \int_A \frac12\ dx & \text{if }A\subseteq[0,1), \\[8pt] \Pr(X=1) = 1/6, \\[8pt] \Pr(X=2) = 1/4, \\[8pt] \Pr(X=3) = 1/12. \end{cases} $$ The expected value is then $$ \int_0^1 x\frac12\, dx + 1\cdot\frac16 + 2\cdot\frac14+3\cdot\frac{1}{12}. $$ This is somewhat pedestrian, but it has this advantage over Robert Israel's answer: the method still works even for random variables that can be negative.