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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 $P(1\le\ X<3)$.

Math Major
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1 Answers1

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Because $F(x)$ is continuous, there is no difference between probabilities defined by strong or weak inequalities. And $P(1\leq X<3)$ is simply $F(3)-F(1)$.

$F(x)$ is the probability, that $X<x$, and $f(x)$ is, at least in the enough regular case, the derivative of $F$. In pracice, you have rather $f(x)$ than $F(x)$ given, because finding an integral is difficult in general.

An example of counting $F(x)$ for $f(x)$ from comment, i.e. $f(x)=cx^2\xi_{[-1,2]}$.

Certainly $$ \int_{-\infty}^{\infty}cx^2\xi_{[-1,2]}\,dx=\int_{-1}^{2}cx^2\,dx=\frac{c}3x^3 \Biggr|_{-1}^{2}=\frac{c}{3}(8+1)=3c, $$ hence $c=\frac13$, beacause the probability must be equal to 1. For $x\in[-1,2]$ we obtain $$ F(x)=\int_{1}^{x}\frac13t^2\,dt=\frac19t^3\Biggr|_{-1}^{x}=\frac{x^3+1}{9}. $$

Przemysław Scherwentke
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  • thank you! but I am a little confused about why F(x) contains the "1-" can you explain that? – Math Major Nov 06 '14 at 15:24
  • or in another example, f(x)=0 for x<-1 or x>2 and $f(x)=cx^2$ for -1<x<2, i don't get why F(x)=(x^3+1)/9 insetad of x^3/9... – Math Major Nov 06 '14 at 15:26
  • @MathMajor I hope that an example in the extended version of my answer will be useful for you. – Przemysław Scherwentke Nov 06 '14 at 15:41
  • so basically you plug in the lower bound to the solution you obtain after you integrate? – Math Major Nov 06 '14 at 15:51
  • @MathMajor Yes, such case as this. But if, say, the density would be determined by different formulas on some intervals, the solution would look different. In this case it is simply an integral from $-\infty$, but up to $x=-1$ it remains equal to zero. – Przemysław Scherwentke Nov 06 '14 at 15:56