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We are given that the probability density function (PDF) of $X$ is

$$f_X(x):=\begin{cases} cx^2& |x|\leq1\\ 0&\text{otherwise}\end{cases}.$$

Now, I find first the constant $c.$ What I got is $3/2$ (I don't know if this is correct.)

Then I tried to solve for $E(X).$ What I got is $0.$

Is this possible or is the value of my constant wrong?

user829347
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X's
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  • "Is [an expected value of zero] possible?" Yes, of course, and they come up frequently in examples. Games whose payout is modeled as a random variable are called "fair games" if the expected value of the payout is zero (cost included). For a simple example, if you flip a fair coin and it lands head you gain a dollar. If it lands tails you lose a dollar. The expected value of the amount you gain or lose is zero here. – JMoravitz Jan 24 '22 at 17:17
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    $E(X)$ is obviously zero because $f_X(x)$ is even function. This fact does not depend on value of $c$ constant. – Ivan Kaznacheyeu Jan 24 '22 at 17:22
  • Integral from -1 to 1 of $f_X(x)$ is $\frac{2c}{3}=1 \Rightarrow c=\frac{3}{2}$ . – Ivan Kaznacheyeu Jan 24 '22 at 17:24
  • So, if we gonna find P(X greater than or equal to 1/2) - there will be no answer because the integral is divergent? – X's Jan 24 '22 at 18:22
  • Why do you think it's divergent? – mihaild Jan 24 '22 at 21:27

1 Answers1

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  • $E(X)=0$ by symmetry.

  • $\int_{-1}^1 x^2 \, dx = 2 \int_0^1 x^2 \, dx = \frac23$. Hence you are right that $c=\frac32$.

  • We have $P(X \ge \frac12) = \int_\frac12^1 cx^2 \, dx=\frac32 \int_\frac12^1 x^2 \, dx. $ Given a pdf, we can evaluate the probability. Note that the upper limit of the integration is $1$ as the density takes values $0$ for value above $1$.

Siong Thye Goh
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