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Suppose $F$ is a cumulative distribution function of a random variable $X$ distributed in $[0,1]$ defined as follows: $$ F(x)= \begin{cases} ax+b & \text{if } x\leq a, \\ x^2-x+1 & \text{otherwise.} \end{cases}$$

where $a\in \left ( 0,1 \right )$ and $b$ is a real number.

What can you say about the continuity and differentiability in $(0,1)$?

I tried to find the pdf and equate it to $1$. The value of $a$ came out to be $1$ , which is not possible.

Any help will be appreciated.

Jean Marie
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  • I think you have it right. It doesn't say that the function has to be differentiable – jg mr chapb Apr 08 '17 at 19:17
  • It never said it was differentiable. It asked you to talk about the differentiability and continuity. Since I don't know anything about random variables, I'm not going to claim the function does not have to be continuous. – Saketh Malyala Apr 08 '17 at 20:28

1 Answers1

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Condition for the function to be continuous:

$\lim_{x \to a}F(x) = a^2 + b$ where $x < a$

= $\lim_{x \to a}F(x) = a^2 - a + 1$ where $x > a$

So $ b = a-1$ to be continuous.

Condition for the function to be differentiable:

$\lim_{x \to a}F'(x) = a$ where $x < a$

$=\lim_{x \to a}F'(x) = 2a-1$ where $x > a$

so $a = 1$ if the function is differentiable. Since this is outside the bounds for $a$, it is not differentiable. It is worth trying to prove that those conditions are equivalent to continuity and differentiability. (Haven't done it myself yet) :p I mean I guess if a function has a continuous derivative and it is continuous within an interval then the function is differentiable

jg mr chapb
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