Joint Probability Distribution Example Problems And Solutions Pdf
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We are currently in the process of editing Probability! If you see any typos, potential edits or changes in this Chapter, please note them here. Thus far, we have largely dealt with marginal distributions.
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- 5.1: Joint Distributions of Discrete Random Variables
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- 5.2: Joint Distributions of Continuous Random Variables
So far, our attention in this lesson has been directed towards the joint probability distribution of two or more discrete random variables. Now, we'll turn our attention to continuous random variables. Along the way, always in the context of continuous random variables, we'll look at formal definitions of joint probability density functions, marginal probability density functions, expectation and independence.
We are currently in the process of editing Probability! If you see any typos, potential edits or changes in this Chapter, please note them here.
Thus far, we have largely dealt with marginal distributions. Thankfully, a lot of these concepts carry the same properties as individual random variables, although they become more complicated when generalized to multiple random variables. Understanding how distributions relate in tandem is a fundamental key to understanding the nature of Statistics. We will also explore a new distribution, the Multinomial a useful extension of the Binomial distribution and touch upon an interesting result with the Poisson distribution.
The best way to begin to frame these topics is to think about marginal, joint and conditional structures in terms of the probabilities that we already know so well. Also recall the importance of thinking conditionally! This should all be familiar to this point, but it is helpful to refresh before applying these concepts to random variables.
This is essentially what we have seen thus far: one random variable by itself. We can now move to joint distributions. Remember that in terms of probabilities of events, the joint probability gives the probability that one event and another event occur.
Largely the same as before: functions of variables, except this time there are two variables in the function instead of one i. So far, this structure is pretty similar to that of basic probability, and the similarity continues with the concept of independence. Formally, what does it mean for two random variables to be independent?
Well, we know that the general relationship between the CDF and PDF is that the latter is the derivative of the former. This principle applies here, except that we have to derive with respect to two variables or, if you have more variables, derive with respect to all of them to get the Joint PDF from the CDF.
Perhaps more important in practice is getting the marginal distribution from the joint distribution. However, often the random variables will not be independent, and another method is needed to recover the marginal PDFs. To find the marginal PDFs, simply integrate continuous case or sum discrete case out the unwanted variable over its support, which will then leave a function that only includes the the desired variable.
You can watch the full lecture freely available that presents this example here. Say that we are looking at the unit circle i. We can confirm this intuition with some simulation and plots in R. Recall that the joint PDF is just 1, so we should see the same color in the entire circle. From here, we can try a common exercise related to joint distributions: marginalizing out one of the variables and thus recovering one of the marginal distributions. This would be easy if the two were independent; however, does it appear that they are independent?
It actually does not. This is a pretty simple integral, but the one thing we have to be careful about is the bounds. It certainly does: there is simply more mass in the middle of the circle than at the edges.
This interval gets smaller and smaller the further we get from 0 simply more mass in the middle ; more on this below. Does this agree when we actually plug into the marginal PDF? Indeed it does. This is a powerful example of symmetry, which was introduced much earlier in the book but is still relevant amidst these more complex topics.
Hopefully by now the gist of joint distributions is clear. In the end, integrals and derivatives help us to navigate the multivariate landscapes of joint distributions. Be sure to always be wary of independence of random variables, just as we are wary of assuming independence for simple probabilities! Another common hazard is working with the bounds when integrating from the joint to the marginal.
We now move from joint to conditional distributions. Recall that a conditional probability is the probability that an event occurs given that another event occurred. Conditional distributions have the same property: it is the distribution of a random variable given that another random variable has crystallized to a specific value.
Can we write this in a different way? Recall the main rule of conditional probability:. Putting it together:. We can reflect on this conditional PDF to help build some intuition. Does that make sense? Hopefully this not only helped in figuring out how to find the conditional distribution but for thinking about how it actually works. You may find it helpful to draw this out on an actual graph if you are still struggling with the visual.
We should see that the density is uniform. Hopefully this provided an enlightening discussion of joint and conditional distributions. Refer back to all of the properties of a joint distribution to build more intuition, and remember that the conditional distribution can be found with the joint and the marginal similarly to how we use joint and marginal probabilities to find conditional probabilities. Now that we are masters of joint distributions - multivariate extensions of marginal distributions - we need an extension of LoTUS as well.
Like its one-dimensional partner, 2-D LoTUS is used in finding expectations of random variables and, eventually, expectations of functions of random variables. The expectation is found in a very similar way: recall how, using LoTUS, we just multiplied the function that we were trying to find the expectation of by the PDF and integrated?
In the two dimensional case, we simply multiply the function of both variables by the joint PDF and integrate over the supports of both variables. Here, we integrate from negative infinity to infinity to be as general as possible; of course, in an actual example, we would only integrate over the support, which might be less extreme than the infinities in the bounds here however, it never hurts to integrate over all real numbers, since if the PDF is defined correctly, any value that is not in the support has no density and will simply return a value of 0.
Standard Uniform random variables. Before we start, do you have any intuition on what this should be? We know marginally that both r. This is by no means a rigorous way to answer the problem, just a way to build intuition to see if our solution matches when we actually perform the calculation. First, we need the joint PDF; thankfully, the two random variables are independent we are given that they are i.
We then need to multiply this simple Joint PDF by the function of the two variables and integrate over the bounds. We write:. Many times, our intuition is not be matched by the eventual result, so enjoy this intuitive result that is also actually correct!
We can check our result very quickly in R, by generating values from two independent standard uniforms and finding the mean of their product. We saw that we arrived at the intuitive answer the mean of the products was the product of the means , but be sure not to over-extend this result.
The above seems like a fast way to the Expectation of the product of random variables: just multiply the Expectation of both random variables. However, while the intuition was enticing in this case, intuition is not an actual proof. This will not only solidify your understanding but will get us a pretty valuable result that we can use in the future.
We can thus re-write:. So, we can finally write:. Clearly, the dependent case requires a bit more work than just multiplying expectations!
How would you find the expected squared difference between two Standard Uniforms? Clearly, this tool can apply to many different types of problems. In fact, it fits in very well with the major underlying theme of this chapter: so far we have looked at concepts developed earlier on, generalized to higher dimensions multiple marginal distributions become joint distributions, one dimensional LoTUS becomes two dimensional, etc.
The Multinomial is a higher dimensional Binomial. Of course, by this point, we are familiar with the Binomial and are quite comfortable with this distribution. You can probably see where the generalization comes in: with the Binomial, we only had two outcomes, success or failure. This is a helpful paradigm for the Multinomial because each object can only be placed in 1 box, which is a key part of the distribution.
The Geometric waits for just 1 success, while the Negative Binomial generalizes to multiple successes. If all of this notation is jarring you, just think again that we are essentially working with a Binomial, but instead of just success or failure we have more than 2 different outcomes. That is, in our notation:. Can you envision why this makes sense?
In that example, we would write the probability as:. Does this make sense? You will find that this is in fact a simple application of the probability and counting techniques you are already very familiar with.
We also know that since all of the trials are independent, we can multiply the probabilities, and thus we get this expression. However, this term only gives the probability of one possible permutation i. Well, consider the number of ways to order How do we correct this? We can use rmultinom to generate random values like rbinom and dmultinom to work with the density like dbinom. We just have to be sure to enter the number of boxes possible outcomes we want, the number of trials we want to have for each experiment and the number of experiments we want to run.
The length of this vector indicates how many outcomes we want our Multinomial random variable to have; here, we have a vector of length 4, meaning the Multinomial has 4 possible outcomes the first possible outcome has probability.
You can see how the output matches the arguments we entered: we have a 4x10 matrix, where every column represents a single experiment. Consider the first column. This is saying that on the first run of the experiment, we put 2 balls in box 3 i.
The columns will always sum to 3, since it is counting the number of balls we put in the boxes. So, these are the general characteristics of the Multinomial, but, given that it is a multivariate distribution remember, this just means a joint distribution of multiple random variables there are interesting extensions that we can tack on.
That makes a lot of intuitive sense, and there are a of couple ways to think about this result. Consider first the tried and true example of rolling the fair die 6 times.
5.1: Joint Distributions of Discrete Random Variables
In this chapter we consider two or more random variables defined on the same sample space and discuss how to model the probability distribution of the random variables jointly. We will begin with the discrete case by looking at the joint probability mass function for two discrete random variables. Note that conditions 1 and 2 in Definition 5. Consider again the probability experiment of Example 3. Given the joint pmf, we can now find the marginal pmf's. We now look at taking the expectation of jointly distributed discrete random variables.
Did you know that the properties for joint continuous random variables are very similar to discrete random variables, with the only difference is between using sigma and integrals? As we learned in our previous lesson, there are times when it is desirable to record the outcomes of random variables simultaneously. So, if X and Y are two random variables, then the probability of their simultaneous occurrence can be represented as a Joint Probability Distribution or Bivariate Probability Distribution. Well, it has everything to do with what is the difference between discrete and continuous. By definition, a discrete random variable contains a set of data where values are distinct and separate i.
Having considered the discrete case, we now look at joint distributions for continuous random variables. The first two conditions in Definition 5. The third condition indicates how to use a joint pdf to calculate probabilities. As an example of applying the third condition in Definition 5. Suppose a radioactive particle is contained in a unit square. Radioactive particles follow completely random behavior, meaning that the particle's location should be uniformly distributed over the unit square.
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5.2: Joint Distributions of Continuous Random Variables
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