0
$\begingroup$

I have a gamma distribution with the following pdf: $$ f(x) = \frac{1}{4} xe^{-0.5x}, x > 0$$

I am trying to determine the shape of the graph without plotting it. I am given a hint ot consider the mean and standard deviation.

I have calculated the mean is $\mu = 4$ and standard deviation is $\sigma = 2\sqrt{2}$, so the standard deviation is large relative to its mean, but I am unable to tell what this mean exactly. Is it positively/negatively skewed based of this?

1 Answers 1

0

The large-$x$ limit is dominated by the exponential term. and extends out infinitely far. (The distribution is bounded by $x = 0$ at the left.) As such, the distribution must be skewed positively. So you know (without graphing):

  • $f(x=0)=0$
  • ${\rm Mean}[f(x)] = 4$
  • There is a single maximum of $f(x)$ (and no inflection points)
  • The distribution tails off to infinity
  • Hence the distribution is skewed positively
  • The standard deviation is $\sigma = 2 \sqrt{2}$
  • $\int\limits_{x=0}^\infty f(x)\ dx = 1$ (of course)

Isn't this enough? Just sketch $f(x)$ from this knowledge and you'll be very close.

  • 0
    I can see it is positive skewed, but my question was not as much as how would I go about sketching it, but more so why the standard deviation and mean in this case can imply +ve skewness? The solution just mentions: "standard deviation is large relative to mean and x>0 so +ve skewed" – 2017-02-21
  • 0
    Because the distribution is bound at $x=0$, has a peak and mean near $x=4$ and extends *to infinity* at the right. By the way: Why *not* merely plot the distribution? What is wrong with that? – 2017-02-21
  • 0
    There is nothing wrong with it per say, but I do not understand why standard deviation can determine its shape – 2017-02-21
  • 1
    The large variance tells you there must be probability relatively far from the mean, and it can't be to the left of 0, so it must far from the mean in a positive direction. – 2017-02-21