If my table is $(KV)^2$ where $K$ is the number of states and $V$ is the number of variables. Thus, I have the conditional probability for any 2 variables. Is it possible to find the joint distribution for any of the variables?
Solved – If I have a table of conditional probability values can I find any joint distribution value
conditional probability
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Those distributions you call "marginal" are not marginal. They are conditional distributions because you wrote $x \mid y$. The marginal distribution of $X$, for example, is necessarily independent of the value of $Y$.
To see how the conditional distribution is gamma, all you have to do is write $$f_{X \mid Y}(x) = \frac{f_{X,Y}(x,y)}{f_Y(y)} \propto f_{X,Y}(x,y).$$ That is to say, the conditional distribution is proportional to the joint distribution, appropriately normalized. So we have $$f_{X \mid Y}(x) \propto x^2 e^{-x(y^2+4)},$$ completely ignoring any factors that are not functions of $x$. Then we recognize that the gamma distribution has density $$f_S(s) \propto s^{a-1} e^{-bs},$$ so the choice of shape $a = 3$ and rate $b = y^2+4$ demonstrates that the conditional distribution $X \mid Y \sim \operatorname{Gamma}(a = 3, b = y^2+4)$.
The conditional distribution of $Y \mid X$ is done in a similar fashion. Just ignore constants of proportionality: $$f_{Y \mid X}(y) \propto e^{-(x+1)y^2+2y},$$ but this one requires us to complete the square to get it to look like a normal density: $$-(x+1)y^2 + 2y = (x+1)\left(-\left(y - \tfrac{1}{x+1}\right)^2 \right) + \tfrac{1}{x+1},$$ and after exponentiating and removing the $e^{1/(x+1)}$ factor, comparing this against $$f_W(w) \propto e^{-(w-\mu)^2/(2\sigma^2)},$$ we see that we have a normal density with mean $\mu = 1/(x+1)$ and variance $\sigma^2 = 1/(2(x+1))$.
Now, if you wanted the marginal distributions, you would need to integrate: $$f_X(x) = \int_{y=-\infty}^\infty f_{X,Y}(x,y) \, dy,$$ for example. And as you can see, this expression will not be a function of $Y$. The difference is that if I simulated realizations of ordered pairs $(X_i, Y_i)$ from the joint distribution, the marginal density for $X$ would be what you would see if I only told you the values of $X_i$. The conditional distribution of $X$ given $Y = y$ would be what you would see if I only gave you the $X_i$ for which the corresponding $Y_i$ was equal to $y$.
Well I think the term "conditional probability" or "conditional probability distribution" can both extend to two or more variables (correct me if I am wrong). For example, let us suppose we have $X,Y,Z$ are i.i.d random variables uniformly distributed on (0,1) for simplicity. Then we are required to find the conditional probability of $P(X \ge YZ|Y>\frac{1}{2})$. This should be a valid question about finding conditional probability.
Then,
$$P(X \ge YZ|Y>\frac{1}{2}) = \frac{P(X \ge YZ,Y>\frac{1}{2})}{P(Y>\frac{1}{2})} = \frac{\int_0^1\int_\frac{1}{2}^1\int_{yz}^1 1 dxdydz}{\frac{1}{2}}$$ $$= 2 \int_0^1\int_\frac{1}{2}^1(1-yz)dydz = 2\int_0^1 \Big(\frac{1}{2}-\frac{3z}{8}\Big)dz=2 \Big( \frac{1}{2}-\frac{3}{16}\Big)=\frac{5}{8}$$
Similarly, $$P(X \ge YZ|Y \le\frac{1}{2})= \frac{\int_0^1\int_0^\frac{1}{2}\int_{yz}^1 1 dxdydz}{\frac{1}{2}} = \frac{7}{8}$$ , as you can verify.
In a nutshell, the concept of conditional probability is not only valid on single variable case.
Best Answer
If the situation is as given below, the table represents your joint probability distributions $P(x,y)$:
Conditionals are normalised rows or columns, i.e. in the table given $P(y | x=1) = P(y, x=1) / P(x=1)$.
So, for example $P(y=1|x=1) = 0.8$.
If you have the conditional probability table to start with, you can acquire the joint probability table provided that you have the marginal probabilities (row and column sums) per each variable state.