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

In summary: If you would like, I can provide a few examples of RV's with Bernoulli distributions that you could look up. However, the accuracy of my posts is not really something I would base my estimation off of.
  • #1
Dustinsfl
2,281
5
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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  • #2
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.
 
  • #3
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 $Y=1-X$

.
 
  • #4
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

$\chi$ $\sigma$
 
  • #5
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.
 

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

What does "Y=-X" mean in the context of the problem?

In this problem, "Y=-X" means that the value of Y is equal to the opposite of X. In other words, if X is 1, Y will be -1. If X is 0, Y will be 0. This relationship is important in solving the mystery.

What does "X ~ Ber(1/4)" indicate in the problem?

This notation means that X follows a Bernoulli distribution with a probability of success (or the value of X being 1) of 1/4. This means that there is a 1/4 chance that X will be 1, and a 3/4 chance that X will be 0.

How do you solve the mystery using the given information?

To solve the mystery, we need to use the given information to find the value of X and Y. We know that Y=-X, and that X follows a Bernoulli distribution with a probability of 1/4. This means that we can use the Bernoulli formula to find the probability of Y being equal to a certain value, which will help us solve the mystery.

Can you provide an example of how to use the Bernoulli formula in this problem?

For example, if we want to find the probability of Y being equal to 1, we can use the Bernoulli formula: P(Y=1) = (1/4)^1 * (3/4)^0 = 1/4. This means that there is a 1/4 chance that Y will be equal to 1. Similarly, we can find the probability of Y being equal to 0 by plugging in the values: P(Y=0) = (1/4)^0 * (3/4)^1 = 3/4.

Are there any other methods for solving the mystery?

Yes, there are other methods for solving the mystery. One method is to use a table or graph to visualize the possible outcomes of X and Y. Another method is to use algebra to solve for the values of X and Y. The method chosen may depend on the individual's personal preference and the complexity of the problem.

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