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I'm having trouble understanding the circled part of this picture below. I see that they set $F(x) = 0$ to solve for $C$, but why were they able to plug in $1$ as opposed to any other number? it seems like a $2$ or $3$ would lead to a different value of $C$.

Also why is $F(x) = 1$ when $x \ge 4$? Is this because of some property of antiderivatives I'm forgetting?

I tried looking through my notes and finding similar examples online, but either all problems are different or they don't really explain why these two steps are taken or why they are allowed to happen.

pdf function problem

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    $\begingroup$ For the bottom right most circle: The cumulative distribution function (cdf) is really just what the name implies - a function that accumulates 'stuff' from the distribution. That 'stuff' we're talking about is the probability of your random variable taking a particular value. The reason why $F(X)$ takes on the value 1 for $x \geq 4$ is really just because it has 'accumulated' a total area of 1 units just as it crosses over the vertical line $x = 4$. $\endgroup$ – Retty Apr 28 '17 at 5:21
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The probability distribution you consider puts no mass outside of $[1,4]$, hence $F(1) = \int_{-\infty}^1 f =0$. Once this is fixed, it imposes $C=-1$. More over, for $x\geq 4$, $1-F(x) = \int_x^\infty f = \int_x^\infty 0 = 0$. Therefore, $F(x)=1$ when $x\geq 4$. In a general setting, if a probability distribution puts no mass outside of $[a,b]$, then $F$ is going to be $0$ on $]-\infty,a[$ and $1$ on $]b,+\infty[$.

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  • $\begingroup$ I'm still not understanding why they are able to just solve for C using F(1) $\endgroup$ – 2316354654 Apr 28 '17 at 5:19
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    $\begingroup$ An antiderivative of a function is defined up to a constant. This constant can be determined when you know that the antiderivative you are looking for takes a special value at a certain point: here, you are looking for the antiderivative of $f$ which is $0$ in $1$. It determines $C$. $\endgroup$ – Vincent Apr 28 '17 at 5:22
  • $\begingroup$ but why do we choose 1 instead of 4 or some number in between? $\endgroup$ – 2316354654 Apr 28 '17 at 5:32
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    $\begingroup$ @2316354654. They simply choose $F(1)$ because the cumulative probability up to that point it 0. $\endgroup$ – Retty Apr 28 '17 at 5:34
  • $\begingroup$ thanks, I think I understand all of it now. $\endgroup$ – 2316354654 Apr 28 '17 at 5:37

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