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I am trying to understand this picture from Statquest

Likelihood of p = 0.25

in the light of the Wikipedia statement that Likelihood

describes the joint probability of the observed data as a function of the parameters of the chosen statistical model.

Am I right thinking "joint probability" does not apply here since we only have 1 dimension, being X?

I am also confused here about what it means to have

a likelihood of p = 0.25 when p = 0.5

surely p cant be both?

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    $\begingroup$ It is a joint probability over all observations. For the image, it is the joint probability of 4 choosing orange and 3 choosing grape. That joint probability varies with possible values of p: 0.273 for p = 0.5, and 0.058 for p = 0.25. $\endgroup$
    – jblood94
    Commented Mar 11, 2022 at 20:27
  • $\begingroup$ Thank you. So are there 7 observations here? Does "joint" refer to grape and orange (the 2 possibilities) or the number of cans? $\endgroup$
    – Kirsten
    Commented Mar 11, 2022 at 21:19
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    $\begingroup$ Yes, there are 7 observations in the graphic. Since one would not be concerned with the specific sequence in which the observations were made, the 7 observations are summarized as "4 chose orange, and 3 chose grape", or, equivalently, "4 out of 7 chose orange over grape". For a given value for p, the joint probability of those 7 observations is given by the binomial distribution with x = 4 and n = 7. $\endgroup$
    – jblood94
    Commented Mar 13, 2022 at 18:43

1 Answer 1

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The likelihood function $L(p; n, x)$ is $L(p; n, x) = \binom{n}{x} p^x(1-p)^{n-x}$ which, regarded as a function of $p$, has value $0.058$ when $n$ and $x$ happen to equal $7$ and $4$ respectively, that is, $L(0.25 \mid 7, 4) = 0.058$ as claimed in the picture you have included. The statement that puzzles you is not what you quote from the picture: it is a mangling of what is actually said in the picture, viz.,

"is less than $0.273$, the likelihood when $p=0.5$."

What the statement is claiming is that $L(0.25; 7, 4) = 0.058$ is a smaller number than $L(0.5; 7, 4) = 0.273$ which everyone will agree with, I hope. It does not say "a likelihood of $p=0.25$ when $p=0.5$" as you think it does.

In short, inferring (and claiming in print advertisements and TV commercials) that $3$ of $4$ dentists recommend Grape Fanta over Orange Fanta when recommending a Fanta flavor to their patients who drink Fanta when the evidence is that $4$ out of $7$ dentists actually surveyed chose Orange Fanta over Grape Fanta is a serious mis-representation of the data and likely to bring condemnation from the readers of stats.SE and maybe even the Advertising Council and the FTC. As the picture says, claiming that dentists are actually equally divided between Orange and Grade Fanta is better since $L(0.5) > L(0.25)$.

Exercise: Show that $L(p; 7,4)$, regarded as a function of $p \in [0,1]$, attains maximum value at $p = \frac 47 = 0.571428\ldots$

Turning to "Likelihood describes the joint probability of the observed data as a function of the parameters of the chosen statistical model"

the joint probability here is the probability that $4$ of the $7$ dentists preferred Orange Fanta while the other three preferred Grape Fanta. This joint probability is the sum of the probabilities of the $35 = \binom{7}{4}$ vectors of $7$ possible answers ($7$-tuples if you prefer that name over vectors) that might be obtained when $4$ out of $7$ dentists say that they prefer Orange Fanta over Grape Fanta and $3$ have the opposite preference. The possible vectors of $7$ answers are \begin{align} &(O,O,O,O, G, G, G)\\ &(O,O,O,G, O, G, G)\\ &(O,O,O,G, G, O, G)\\ &(O,O,O,G, G, G, O)\\ &(O,O,G,O, O, G, G)\\ &(O,O,G,O, G, O, G)\\ &(O,O,G,O, G, G, O)\\ &\text{and so on till the $35$-th item}\\ &(G,G,G, O, O, O, O) \end{align} Each vector has probability $p^4(1-p)^3$ (hidden assumption: the dentist preferences are independent and each dentist has probability $p$ of preferring Orange Fanta over Grape Fanta) giving that the _joint probability that $4$ of $7$ dentists prefer Orange Fanta over Grape Fanta is $\binom{7}{4}p^4(1-p)^3 = 35p^4(1-p)^3$. How come we can just add up these $35$ probabilities? Well, if one specific vector of answers is observed, we are automatically excluding all other $34$ vectors of answers as possible: these are disjoint (a.k.a. mutually exclusive) events and so the probability that some one of these $35$ vectors (we don't care which one) was observed (i.e. some $4$ out of the $7$ dentists asked preferred Orange Fanta over Grape Fanta) is $\binom{7}{4}p^4(1-p)^3 = 35p^4(1-p)^3$. For each $p \in [0,1]$, the likelihood function (regarded as a function of $p$, with $n$ and $x$ as parameters), has value equal to the joint probability of the observations given the parameters. Thus, $L(p; 7, 4) = \binom{7}{4}p^4(1-p)^3$.

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  • $\begingroup$ Thank you for clearing my confusion about the wording. I also asked about how joint probability applies here. Should that be a separate question? $\endgroup$
    – Kirsten
    Commented Mar 11, 2022 at 18:17
  • $\begingroup$ Thank you for the joint probability update. However looking at Wikipedia, a joint probability distribution involves two random variables. Surely we have only one here? that is X? Only X is mentioned in the Likelihood equation? Or is the other RV implied somehow as N-X? From your answer I guess it must be ... but surely N-X is not an RV indepentant of X? en.wikipedia.org/wiki/Joint_probability_distribution $\endgroup$
    – Kirsten
    Commented Mar 12, 2022 at 6:38
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    $\begingroup$ @Kirsten Repeat as often as needed till you have memorized it: "X is not a random variable". There are 7 independent random variables representing the 7 answers; these random variables take on value Orange with (unknown) probability $p$ and value Grape with probability $1-p$. 4 of the answers are known to have value Orange and 3 to have value Grape. Before asking for the choices, we know that the joint probability that 4 of these 7 random variables will have value Orange and 3 will have value Grape is $\binom 74p^4(1-p)^3$. Repeat again: "X is not a random variable". $\endgroup$ Commented Mar 12, 2022 at 14:44
  • $\begingroup$ Thank you, but why is X not an RV? It looks like a "numerical description of the outcome of a statistical experiment." (though I now appreciate that the independent RVs in this problem must be the binary drink outcome ) britannica.com/science/statistics/… $\endgroup$
    – Kirsten
    Commented Mar 13, 2022 at 12:10
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    $\begingroup$ @Kirsten Well, yes, $X$ is indeed a binomial random variable, but you are insisting on finding a joint distribution of $X$ and some other random variable. The outcomes of the experiment (the data) are the 7 answers and the joint distributions that you ought to be seeking are the joint distribution of the 7 answers, and not the joint distribution with something else. $\endgroup$ Commented Mar 13, 2022 at 20:44

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