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$begingroup$Lads, I need some insight here.
Given our $f(x) = sqrt{frac2pi}e^{-(x^2/2)}$ where x is non-negative, I have to prove that $f(x)$ is a valid pdf.
My work done so far:I've tried to pull out the constant, $sqrt{frac2pi}$, out of the integral, to make the integral look like the normal distribution and I keep getting 2 instead of 1, which is not what I want because I have to get a 1 to prove that it's a valid pdf. What am I missing here?
Ddlc mod a brand new day. Apologies for my poor symbols.
George LawBrowse other questions tagged probability-distributions or ask your own question.
$begingroup$So in order to prove that $f(x)=1/(x+1)^2$ $for$ $x>=0$ and $0$ $otherwise$ is a probability density function, it needs to satisfy the 'properties' of a PDF which are:
1) $0<= f(x)$
2) $int f(x) = 1$
3) $P(A) = P(a<=X<=b) = int_A f(x)$ $dx$
How To Prove It Solutions
to prove each:
1) Can I just show that since $lim_{xtoinfty} f(x)=0$ and $f(0)$=1 and it's an obvious decreasing function (by graphing?) then $0<= f(x)$
2) $int f(x) = int_0^infty 1/(x+)^2$ = -1/(x+1) = 0 - (-1) = 1
3) this is where I'm stuck. I don't quite know how to prove this property. I tried the following:
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$P(A) = P(a<=X<=b) = P(X<=b) - P(X<a)= 1/(a+1) - 1/(b+1)$
$and$
$int_a^b f(x)$ $dx$ $= 1/(a+1) - 1/(b+1)$
is this the right way to do it?
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