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This is a question I want to solve:

The random variable $X$ has cdf: $$ F_X(x) = \begin{cases} \begin{align} &0 &&x <0\\ &0.5 + c\sin^2\left(\frac{\pi x}{2}\right) &0 \leq\; &x \leq 1\\ &1 &&x>1 \end{align} \end{cases} $$

(a) What values can $c$ assume?

(b) Plot the cdf.

(c) Find $P[X > 0]$.

I assume that $x$ is between $0$ and $1$ so $$ 0.5 + c \sin^2(\pi\times x/2) = 0 $$

then $x=1$ and $$ c \sin^2(\pi\times 1/2) = -0.5 $$

since $c = -0.5$

Is that correct and I need help with the nother parts please.

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You have no reason to assume that $0.5 + c \sin^2(\pi\times x/2) = 0$. In fact the definition of $F$ tells you that $$F_X(0) = 0.5 + c \sin^2(\pi\cdot 0/2) = 0.5 \neq 0,$$ so your assumption is false for at least one $x$ in the interval $0 \leq x \leq 1$.

There is also no reason to assume that $c = -0.5$. In fact it cannot be $-0.5.$

I would suggest looking carefully at what happens around $x=1.$ It is possible for a cdf to be discontinuous, but only some kinds of discontinuity are possible.

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Hint: A CDF is monotonic increasing and right continuous. However, $\sin^2(x\pi/2)\in [0,1]$

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The nature of a cumulative distribution function is that it must be càdlàg, monotonically non-decreasing, and take values between 0 and 1 (inclusive). It does not have to be either continuous or discrete, but may be both.

  • Part 1: Being that the distribution may not take values greater than one, and must be non-decreasing, what can the maximum of $c$ be if we know it is equal to $0.5 + c\sin^2\left(\frac{\pi x}{2}\right)$? What is $\sin\left(\frac{\pi}{2}\right)$? Similarly, what is the minimum it can be, remembering that it must be non-decreasing?
  • Part 2: Being that $c$ is a constant, there are really only two kinds of shapes it can have, so once you figure out part 1, this isn't so difficult
  • Part 3: Make sure you recognize those inequalities which are strict, and those that aren't, and what the difference must mean.
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I think the CDF given in the question is wrong as in your problem it is neither a continuous CDF nor a discrete one as LHL (left hand limit) approaching 0 and RHL (right hand limit) approaching 1 do not converge to the same value. Which in this case is not true for F(x) at $0\leq x \leq 1$.

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  • $\begingroup$ The distribution is neither continuous nor discrete. It is possible for such a distribution to have a cdf that is not continuous. $\endgroup$ – David K Feb 6 '15 at 5:03
  • $\begingroup$ Consider that there may be mass points for some values, and continuous probability for others. $\endgroup$ – Avraham Feb 6 '15 at 5:14

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