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The density function for a random variable X is given in terms of a constant $c$. Find the value of $c$. What is the corresponding distribution function? Sketch both the density and the distribution functions. Finally, find the probabilities.

5.1 $f(x)=0$ for $x<0$ and $f(x)=\frac{c}{(x+1)^4}$ for $0>x$

  • $P(X>4)$
  • $P(X<2)$
  • $P(1\le\ X<3)$

I get $c=3$ by taking the integral, and then I am a little confused as to why $F(x)=1-\frac{1}{(x+1)^3}$

Is that because in solving for $c$, I ended up with $1=\frac{c}{3(x+1)^3}$, so substituting c in I get $1=\frac{1}{(x+1)^3}$?

And, then

$P(X>4)=1-P(X\le\ 4)$ $=1-(1-{1}{(1+4)^3})$ $=1/125$

$P(X<2)=1-{1}{(1+2)^3}$ $=26/27$

I guess I am a little confused about what $F(x)$ and $f(x)$ actually MEAN? And thus that leads to my confusion about what the differences between $>$ and $<$ and greater than or equal to are, etc etc in terms of the equations. The ones above I solved by comparing to the book but I still don't understand what they mean really.

So as a result, I don't really get how to solve $P(1\le\ X<3)$.

5.2 $f(x)=ce^x$ for $x<0$ and $f(x)=ce^{-x}$ for $0>x$

5.3 $f(x)=0$ for $x<-1$ or $x>2$ and $f(x)=cx^2$ for $-1<x<2$

Henry
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Math Major
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    Do you have a typo ? Is this what you wanted to post ?:$f(x)=\frac{c}{(x+1)^4}$ for $0<x$. The exponent only for the denominator and the inequality sign other way round. – callculus42 Nov 06 '14 at 06:11

1 Answers1

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In 5.1, you presumably intended $f(x)=\dfrac{c}{(x+1)^4}$ for $x \ge 0$, as callculus commented.

You then have $F(x)= \Pr(X \le x)=\displaystyle \int_0^x f(y)\,dy = c\left(-\frac{1}{3(x+1)^3}+\frac{1}{3}\right)$ for $x \ge 0$. Rather more easily, $F(x)=0$ for $x \le 0$

For this to tend to $1$ as $n\to \infty$, you will need $c=3$ so $F(x)=1-\dfrac{1}{(x+1)^3}$.

You then get

  • $\Pr(X\gt 4)=1-F(4)=1-\left(1-\frac{1}{125}\right)=\frac{1}{125}$
  • $\Pr(X \lt 2)= \Pr(X \le 2) = F(2) = 1-\frac{1}{27}=\frac{26}{27}$
  • $\Pr(1\le X\lt 3)= \Pr(X\lt 3) - \Pr( X\lt 1) = F(3)-F(1) = \left(1-\frac{1}{64} \right)-\left(1-\frac{1}{8} \right)=\frac{7}{64}$

Questions 5.2 and 5.3 can be answered in a similar manner, though in 5.3. you need to be aware that the density is zero above $2$ so $\Pr(X \gt 2)=0$ and you will have to further restrict cases where $F(x)$ is neither $0$ nor $1$.

Henry
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  • This confuses the conditions $x>0$ and $x>1$. As a consequence, $c=3$ as the OP says, not $c=24$ (and the rest of the answer should be likewise corrected). – Did Sep 01 '16 at 11:36
  • @Did - thank you for pointing that out – Henry Sep 01 '16 at 14:16