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For $y = \frac{1}{4}\text{ }x\text{ }e^{-x/2}$, my initial hunch was that it is a normal distribution but I wasn't able to figure out what the mean and variance would be. What is it? What are its mean and variance?

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    $\begingroup$ 1. Is this for some subject? If not, how does the question arise? Either way, it lacks context. 2. Note that hunches are not a good way to make progress in statistics. (It's clearly not normal, since it has an "$x$" out the front of the exponent and lacks any $-x^2$ type term in the exponent -- did you even try to draw it?) 3. Is this meant to be a density function for a continuous variate on the positive half-line, or something else? 4. What distributions do you know about? What search and research have you done on this? $\endgroup$ – Glen_b -Reinstate Monica Jan 29 '17 at 2:31
  • $\begingroup$ sorry i should've been more clear. This is a probability density function for a continuous RV. I derived this from a joint multivariate density function. It's an example from my statistics course and on top of just finding the prob density, the question is asking to identify the distribution of y. $\endgroup$ – Derek Frank Jan 29 '17 at 3:24
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As a previous answer has suggested, it appears the support for this distribution is:

$$x\in[0,\infty)$$

To verify it as a density, we can integrate across the support and the area should be equal to 1:

$$\begin{align} \frac{1}{4}\int_{0}^{\infty}xe^{-x/2}\,dx&=\frac{1}{4}\Big[-2xe^{-x/2}\Big]_{0}^{\infty}+\frac{1}{4}\int_{0}^{\infty}2e^{-x/2}\,dx\\ &=0+\frac{1}{4}\Big[-4e^{-x/2}\Big]_{0}^{\infty}\\ &=1 \end{align}$$

Now, as to what density this is, note that the $\text{Gamma}(\alpha,\beta)$ density has the form:

$$f(x)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$$

If we let $\alpha=2$ and $\beta=0.5$, this leads to:

$$f(x)=\tfrac{1}{4}xe^{-x/2}$$

Thus, your distribution is that of a $\text{Gamma}(2,0.5)$. This verifies the mean and variance calculations:

$$\begin{align} E[Y]&=\frac{\alpha}{\beta}=4\\ \text{Var}(Y)&=\frac{\alpha}{\beta^{2}}=8 \end{align}$$

This also confirms the support we defined above. One way that such a Gamma distribution can arise, for example, is if we have:

$$X_{i}\sim \text{Exp}(\lambda=0.5)$$

where all $X_{i}$ are independent and $$Y=\sum_{i=1}^{2}X_{i}$$

then $$Y\sim\text{Gamma}(2,0.5)$$

As to why you might think it is a normal distribution, you should note that the equation for $y$ does not follow the form of a normal density at all. Further inspection of the plot of $y$ shows that the density is skewed, which should tell you immediately that it cannot be normal.

enter image description here

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Evidently the range of this RV is $(0, \infty)$, because that's how you get the density to integrate to 1.

The mean is $$\frac{1}{4} \int_{0}^{\infty} x^2 {\rm exp}(x/2) dx = 4$$

This is easy to solve using integration by parts.

Through a similar (but more tedious) calculation, you can find that the second moment is

$$\frac{1}{4} \int_{0}^{\infty} x^3 {\rm exp}(x/2) dx = 24$$

which means the variance is $24 - 4^2 = 8$.

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  • $\begingroup$ right thanks for that, but looking back at the graph. it doesnt seem normally distributed $\endgroup$ – Derek Frank Jan 29 '17 at 3:50
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    $\begingroup$ The density function you specified is not that of the normal distribution. $\endgroup$ – gammer Jan 29 '17 at 5:20
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    $\begingroup$ @DerekFrank, For one thing, it's the density of a positive random variable. And, second, it's skewed. So the normal distribution has nothing to do with this. Not sure why you think it does. $\endgroup$ – gammer Jan 29 '17 at 5:22
  • $\begingroup$ In case it is of help... $\frac{1}{4} \int_{0}^{\infty} x^2 {\rm exp}(\color{red}{-}x/2) dx = 4$? $\endgroup$ – Antoni Parellada Feb 4 '17 at 19:49
  • $\begingroup$ @AntoniParellada, Yep. Looks like a typo. $\endgroup$ – gammer Feb 5 '17 at 0:33

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