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When using the likelihood method, plotting the relative likelihood or evidence against the range of possible values for the parameter ($\theta$) being estimated results in a curve. The maximum value will indicate the estimated parameter with the maximum likelihood (most supported by the data), $MLE=argmax_\theta \mathcal{L}(\theta|x)$, and a range can be extracted such that all the $\theta$ values are supported more than $1/8$ (horizontal orange line below) or $1/16$ (horizontal gray line) as compared to the maximum are highlighted (an estimated range).

For example, in the experiment of throwing a die multiple times, and observing the number of heads to estimate the probability of getting heads (bias),

n <- 4 # Number of coin flips
x <- 3 # Number of heads
theta <- seq(from=0, to=1, by = 0.001) # Possible range of probabilities of getting H.
L <- theta^x*(1-theta)^(n-x) # Likelihood of each theta given x.
df<- as.data.frame(cbind(theta,L))
maxL <- max(L)
ML_theta<-df$theta[which(df$L==maxL)]

plot(L/maxL ~ theta, type='l', col='firebrick', lwd=2, xlab="Theta", 
     ylab="Relative likelihood")
lines(range(theta[L/maxL>1/8]), c(1/8,1/8), col='orange', lwd=2)
lines(range(theta[L/maxL>1/16]), c(1/16,1/16), col='slategray', lwd=2)
arrows(x1=ML_theta,x0=ML_theta, y1=0, y0=1, col='blue')

the MLE is $\theta = p(H)=0.75$ if we know that we got $3 \, heads\,1\, tail$, but the possible true probability of getting heads can be different, except for $0$ or $1$, and we may give a range of $\theta$'s with a minimum support of $1/8$, range(theta[L/maxL>1/8])= 0.262 to 0.986, (or similarly, we could do the same for $1/16$, yielding 0.203 to 0.993) relative to the maximum. This makes sense.

The question refers to the wording in the Coursera course (Brian Caffo, "Mathematical Biostatistics Boot Camp"; lecture 6 (slide 15)) I came across this, and the wording that is repeated several times in the recorded lecture:

"Parameter values above the $1/8$ reference line, for example, are such that no other point is more than 8 times better supported given the data."

Is the wording obtruse by necessity or by choice? Is my interpretation above wrong or incomplete, explaining why I'm missing the need for this painful and cryptic wording of the meaning of the plot? What is the lecturer trying to convey by making it so difficult? OK, I admit it, I don't know what he means... Can someone please explain?

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    $\begingroup$ The plot is really a plot of the likelihood function. A "maximum likelihood plot" would presumably consist just of the point (0.75,1). $\endgroup$ – Glen_b -Reinstate Monica Aug 14 '15 at 0:09
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OK... I had to think a bit about this tongue-twister.

In the plot in the OP, the y-axis shows the relative likelihood with a maximum of $1$, because every possible $\theta$ value in the $x$-axis is compared to the maximum likelihood point or value, which will have $1$ in the $y$ axis (it's divided by itself... sort of a normalization procedure).

By drawing a horizontal line along $1/8$ in the $y$ axis we guarantee that NONE of the $x$ axis points inside the interval defined by that line and the intersection with the curve will be MORE than eight times better supported by the data than the endpoints of the line (or segment) - the points of crossover of the line with the curve.

To see this, the question to ask is, are there ANY points within the interval limited by the line $two times$ (for example) better supported by the data? And the answer would be... any point in the interval defined by the horizontal line through $1/4$ (red line in the plot below), because any point along the red segment will have a relative evidence compared to the ends of the $1/8$ line at least $\frac{1/4}{1/8}=2$ times greater.

Yet, we won't find any points WITHIN the $1/8$ confines MORE than eight times better supported - the maximum is the maximum likelihood point ($1$), which is $\frac{1}{1/8}=8$ times better supported, but NOT... MORE than $8$.

So straightforward... In any event, I'm posting it in case someone else gets stuck with this play of words.

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