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Can you explain me, why $P(x)>1$ or why is the value of the function $≈1.2$ ?

$P(x) ≤1$, right?

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    A probability density doesn't have to be less than $1$, as long as it doesn't stay that way for an interval of length $1$. – Ian Apr 15 '18 at 15:48
  • @lan Is this a density function? what is the highest value this function? –  Apr 15 '18 at 15:56
  • @Martin R İs this function not a Probability distribution function? –  Apr 15 '18 at 16:05
  • @Beginner how would we know if it is a density function? It is your question and there are no words to suggest if it is or it isn't. The answers focus around this because it is most likely that that is what you mean. – John Doe Apr 15 '18 at 16:06
  • @Beginner again, we can't tell you what the highest value is if we don't have an equation for the function. Do you know the equation? It may be an equation involving $\alpha,s,m$, judging by the key on the graph. – John Doe Apr 15 '18 at 16:07
  • I was studying Internet resources (related Probability). I wanted to ask because, ou I didnt understand this picture. I am sorry this stupid question.. –  Apr 15 '18 at 16:12
  • No don't worry, it isn't a stupid question and is probably common to people who are learning the topic. I suppose if you included that comment (that you were studying some online resources), it might have been better. No harm done though – John Doe Apr 15 '18 at 17:18

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The figure in the question appears to show graphs of several probability distribution functions. A probability distribution function is not a probability.

In order to convert a probability distribution function into a probability, you can integrate it. The integral can never be greater than $1,$ but the function you are integrating can be greater than $1$ over a small enough interval.

David K
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  • Thank you for answer. What is the highest value this function? –  Apr 15 '18 at 16:03
  • As someone has commented already, since you have not given formulas for the graphs we cannot say for sure what the maximum value is. All we can do is guess. – David K Apr 15 '18 at 16:12
  • I understood. Thank you very much. –  Apr 15 '18 at 16:14
  • As DavidK says, for a continuous distribution like this you need to integrate to find probability values, i.e. probability corresponds to area. In this analogy your question is like asking "If I've got 1 square inch of material, how come I can re-shape it into a shape that is longer than 1 inch?". – JonathanZ Apr 15 '18 at 16:19
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What you see is not $p(x)$, it is $pdf(x),$ which stand for probability density function.

With continuous distributions $p(x)=0$ at every $x$, but $p(a<x<b)$ is the area under the graph of $y=pdf(x)$

Mohammad Riazi-Kermani
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$P(x)\le1$, right?

Wrong - what you are thinking of is the fact that for a discrete probability distribution, all probabilities must be at most $1$, since they have to sum to $1$. This is a continuous distribution, so this restriction no longer holds - the sum being equal to $1$ is represented by an integral: $$\int_\Bbb R P(x)\,\mathrm dx=1$$There are no other restrictions on $P$ other than that it is non-negative.

John Doe
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