MHB Consider the following probability mass function of a random variable x

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The provided probability mass function, $p(x) = p^x (1-p)^{1-x}$ for $x=0.1$ and $0$ otherwise, is not valid as it does not sum to one. Since $x$ only takes the value of 0.1, $p(0.1)$ should equal 1, which is not achievable with the given function. Consequently, the mean of the random variable $X$ is 0.1, and the variance is 0, as there is no deviation from the mean. This indicates that the function fails to meet the criteria for a proper probability mass function.
dylbester
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$p(x) ={p}^{x}*{(1-p)}^{1-x}$
for $x=0.1$
$0$ otherwise

Where $p$ is such that $0<=p<=1$

Question:
Find the mean and variance of $X
 
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dylbester said:
$p(x) ={p}^{x}*{(1-p)}^{1-x}$
for $x=0.1$
$0$ otherwise

Where $p$ is such that $0<=p<=1$

Question:
Find the mean and variance of $X$

Hi dylbester! Welcome to MHB! (Smile)

That doesn't seem to be a proper probability mass function.
The sum of $p(x)$ for all values of $x$ is supposed to be $1$.
Since there is only one value for $x$, which is $0.1$, that should mean that $p(0.1)=1$, but that is not the case for any choice for $p$.

Anyway, if there is only one value for $x$, the mean will have to be that value.
And the deviation of that mean can only be $0$, so the variance will then have to be $0$.
 
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