Y=-X if X ~ Ber(1/4): Solving the Mystery

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SUMMARY

The discussion centers on the transformation of random variables, specifically analyzing the relationship between \(Y = -X\) and \(X \sim Ber(1/4)\). Participants concluded that while \(Y\) can be expressed as a distribution, it does not follow a Bernoulli distribution, which is limited to values 0 and 1. Instead, \(Y\) is defined as \(Y \sim \begin{cases} 1 - p, & y = 0\\ p, & y = -1 \end{cases}\) with \(p = 0.25\). The conversation emphasizes the need to understand random variable transformations, particularly in discrete contexts.

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Dustinsfl
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If \(Y = -X\) and \(X\sim Ber(1/4)\), then what is Y?

I know that
\[
X\sim
\begin{cases}
1 - p, & x = 0\\
p, & x = 1
\end{cases}
\]
where \(p = 0.25\) in this case. What is the negative of \(X\) though. It doesn't make any sense making the probabilities negative.
 
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I'm not confident in this answer but I would consider this to also follow a Bernoulli distribution by the following:

$Y\sim
\begin{cases}
1 - p, & y = 0\\
p, & y = -1
\end{cases}$

You should look up random variable transformations and maybe you can find some examples with discrete transforms. The examples that come to mind I've done in the past year have all been for continuous distributions and involve using the CDF.
 
Jameson said:
I'm not confident in this answer but I would consider this to also follow a Bernoulli distribution by the following:

$Y\sim
\begin{cases}
1 - p, & y = 0\\
p, & y = -1
\end{cases}$

You should look up random variable transformations and maybe you can find some examples with discrete transforms. The examples that come to mind I've done in the past year have all been for continuous distributions and involve using the CDF.

That is the correct distribution, but it is not Bernoulli. A RV with a Bernoulli distribution takes only the values 0 or 1.

Given dwsmiths history of accuracy of posting questions I would not be supprised if what he was really asked for was the distribution of $Y=1-X$

.
 
zzephod said:
That is the correct distribution, but it is not Bernoulli. A RV with a Bernoulli distribution takes only the values 0 or 1.

Given dwsmiths history of accuracy of posting questions I would not be supprised if what he was really asked for was the distribution of $Y=1-X$

.

Your observation is not particularly useful from a practical point of view. Let's suppose to have two independent variables X and Y with exponential distribution and we want to find the distribution of the variable Z = X - Y. It is clear that it is necessary to determine the distribution of the variable - Y that will still exponential only instead of y appears -y and its domain will include all the $y \le 0$. Same problem of course if X and Y have Bernoulli distribution ...

Kind regards

$\chi$ $\sigma$
 
zzephod said:
Given dwsmiths history of accuracy of posting questions I would not be supprised if what he was really asked for was the distribution of $Y=1-X$

.

Well, I can tell you are incorrect with your hypothesis.
 

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