How do you sketch the Poisson distribution function: $L(\mu;n)=\frac{\mu^{n}e^{-\mu}}{n!}$? 
When doing a maximum likelihood fit, we often take a ‘Gaussian
  approximation’. This problem works through the case of a measurement from a Poisson
  distribution. The Poisson distribution results in a likelihood for the average number $\mu$, given that $n$ events were observed:$L(\mu;n)=\cfrac{\mu^{n}e^{-\mu}}{n!}$


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Sketch $L(\mu;n)$. It may also be useful to sketch separately the two terms, $\mu^n$ and $e^{-\mu}$, which contribute to the overall shape.

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$L(\mu;n)$ is shown below for the example of $n = 3$. The dashed lines are $\frac{\mu^3}{6}$ and $e^{-\mu}$ separately:

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My question is very simple: How on earth do those dashed lines contribute to the solid black line?
 A: For example, look at where the dotted lines intersect, the height appears to be .3 for both. Thus,
$$.3\cdot .3 = .09$$
which is about the height of the solid line.
Similarly as you examine farther out.
A: They contribute by their pointwise product, though it is not easy to guess, as their graphs are cut above $0.4$. First look at the dot at $(0.8,0.4)$. Take a vertical dive (with a ruler, a sheet of paper), you will see that the corresponding point on the second dashed curve ($u^3/6$) has coordinates $(0.8,0.1)$. On the black curve, $u=0.8$ corresponds approximately to $0.04 \simeq 0.4\times 0.1$.
Same for the crossing on the dashes. They meet approximately at $(1.1,0.28)$, and the black curve passes through $(1.1,0.9)$, where $0.9 \simeq 0.28^2$. Additionally, the second dashed curve is zero at $u=0$, which is consistant with the origin of the black curve.
You can redraw the three curves by yourself, or recheck the values with a pocket calculator. The next figure, drawn with RechnerOnline, provides you with a finer illustration. For instance, you can see how the green and red curves nicely intersect close to $x=1.8$. Indeed, with $x\simeq 1.81712$, $x^3/6\simeq 1$, hence the crossing $y$ is that of $\exp(-1.81712) \simeq 0.16249$.

