A Continuous Multivariate Distribution

In summary, the random variable (x,y) with density f(x,y) = ce^{-(ax+by)} for 0≤y≤x≤1, with given constants a>0 and b>0, has a constant c = ab(a+b)/(e^{-(a+b)}-(a+b)e^{-a}+b). The conditional probability density f_y(y|x) is given by the joint density divided by the marginal density of x. The regression curve of Y on X, or E(Y|X=x), is equal to (e^{-ax}/(e^{-ax}-e^{-(ax+bx)})) * (1-(b+1)e^{-b})/b^2.
  • #1
squenshl
479
4

Homework Statement


The random variable ##(x,y)## has density ##f(x,y) = ce^{-(ax+by)}## for ##0\leq y\leq x\leq 1##, with given constants ##a > 0##, ##b > 0##.
1. Compute the constant ##c##.
2. Find the conditional probability density ##f_y(y|x)##.
3. Compute the regression curve of ##Y## on ##X## i.e. ##E(Y|X = x)##.
4. Sketch this regression curve for ##a = 1##, ##b = 2##.

Homework Equations

The Attempt at a Solution


For 1. my limits of integration for both ##x## and ##y## is ##0## and ##1##. This gives (after doing the double integration) ##c = \frac{ab}{e^{-(a+b)}-e^{-a}-e^{-b}+1}##. Is this right?
For 2. the conditional density is given by the joint density divided by the marginal density of ##x##. I just want to make sure that my ##c## is correct first.
Not sure on 3. and 4.
 
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  • #2
squenshl said:

Homework Statement


The random variable ##(x,y)## has density ##f(x,y) = ce^{-(ax+by)}## for ##0\leq y\leq x\leq 1##, with given constants ##a > 0##, ##b > 0##.
Did you write the conditions on ##x## and ##y## exactly as they were given to you? If so, the limits on your integral and your answer to #1 are wrong.
 
  • #3
vela said:
Did you write the conditions on ##x## and ##y## exactly as they were given to you? If so, the limits on your integral and your answer to #1 are wrong.
I did ##c\int_0^1 \int_0^1 e^{-(ax+by)} \; dx \; dy = 1## and solved for ##c##. If this is wrong, then my problem is how to interpret ##0\leq y\leq x\leq 1##.
 
  • #4
squenshl said:
My problem is how to interpret ##0\leq y\leq x\leq 1##.
You need to take into account that it says ##y \le x##.
 
  • #5
squenshl said:

Homework Statement


The random variable ##(x,y)## has density ##f(x,y) = ce^{-(ax+by)}## for ##0\leq y\leq x\leq 1##, with given constants ##a > 0##, ##b > 0##.
1. Compute the constant ##c##.
2. Find the conditional probability density ##f_y(y|x)##.
3. Compute the regression curve of ##Y## on ##X## i.e. ##E(Y|X = x)##.
4. Sketch this regression curve for ##a = 1##, ##b = 2##.

Homework Equations

The Attempt at a Solution


For 1. my limits of integration for both ##x## and ##y## is ##0## and ##1##. This gives (after doing the double integration) ##c = \frac{ab}{e^{-(a+b)}-e^{-a}-e^{-b}+1}##. Is this right?
For 2. the conditional density is given by the joint density divided by the marginal density of ##x##. I just want to make sure that my ##c## is correct first.
Not sure on 3. and 4.

In problems involving two-dimensional densities, always sketch the correct (x,y)-region before you start trying to do integration. Your error will stare you in the face if you do that.
 
  • #6
Oh ok so in that case I get ##y## from ##0## to ##x## and ##x## from ##0## to ##1##.
 
  • #7
vela said:
You need to take into account that it says ##y \le x##.
squenshl said:
Oh ok so in that case I get ##y## from ##0## to ##x## and ##x## from ##0## to ##1##.
Using these new limits i get ##c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}##
 
  • #8
Using these new limits i get ##c = \frac{ab(a+b)}{e^{-(a+b)}-(a+b)e^{-a}+b}##
 
  • #9
It looks like you made an algebra mistake somewhere. You should have gotten
$$c = \frac{ab(a+b)}{a e^{-(a+b)}-(a+b)e^{-a}+b}.$$
 
  • #10
vela said:
It looks like you made an algebra mistake somewhere. You should have gotten
$$c = \frac{ab(a+b)}{a e^{-(a+b)}-(a+b)e^{-a}+b}.$$
Yes I did get that just forgot to put the ##a## in there haha. Thanks.
 
  • #11
To find the marginal density do I calculate
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{1} e^{-(ax+by)} \; dy?
\end{split}$$
or is it
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{x} e^{-(ax+by)} \; dy?
\end{split}$$
 
  • #12
squenshl said:
To find the marginal density do I calculate
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{1} e^{-(ax+by)} \; dy?
\end{split}$$
or is it
$$\begin{split}
f_X(x) &= \int_{-\infty}^{\infty} f(x,y) \; dy \\
&= c\int_{0}^{x} e^{-(ax+by)} \; dy?
\end{split}$$
Still not sure which it is and why. Please help.
 
  • #13
squenshl said:
Still not sure which it is and why. Please help.
Never mind. Pretty sure it's the second one.
 
  • #14
I'm asked to find the conditional probability function ##f_y(y|x)##. This is
$$\begin{split}
f_y(y|x) &= \frac{f(x,y)}{f_x(x)} \\
&= \frac{ce^{-(ax+by)}}{\frac{c}{b}\left[e^{-ax}-e^{-(ax+bx)}\right]} \\
&= \frac{be^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}}.
\end{split}$$
Just want to make sure I'm good.
I'm also asked to find the regression curve (which is the one I'm struggling with) ##E(Y|X=x)##. I get
$$\begin{split}
E(Y|X=x) &= \int_0^1 y f_y(y|x) \; dy \\
&= \int_0^1 \frac{bye^{-(ax+by)}}{e^{-ax}-e^{-(ax+bx)}} \\
&= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy.
\end{split}$$
We use integration by parts on the integral above. Let ##u=y##, so ##du=dy##. Also ##dv = e^{-by}##, so ##v = -\frac{e^{-by}}{b}##. Hence,
$$\begin{split}
E(Y|X=x) &= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \int_0^1 ye^{-by} \; dy \\
&= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1+\frac{1}{b}\int_0^1 e^{-by} \; dy\right) \\
&= \frac{be^{-ax}}{e^{-ax}-e^{-(ax+bx)}} \left(\left[-\frac{ye^{-by}}{b}\right]_0^1-\left[\frac{e^{-by}}{b^2}\right]_0^1\right) \\
&= \frac{e^{-ax}}{e^{-ax}-e^{-(ax+bx)}}\frac{1-(b+1)e^{-b}}{b^2}.
\end{split}$$
My question is are the limits of integration ##0## to ##1## shown or ##0## to ##x## or even another set of limits?
Thanks again.
 
  • #15
I guess I'm correct then.
 
  • #16
Nope that is wrong.
$$E(Y|X=x) = \int_0^x \frac{bye^{b(x-y)}}{e^{bx}-1} \; dy = \frac{1}{b}-\frac{x}{e^{bx}-1}$$
after several lines of working.
Is this right as my limits of integration are ##0## and ##x##?
 

FAQ: A Continuous Multivariate Distribution

1. What is a continuous multivariate distribution?

A continuous multivariate distribution is a statistical model that describes the probability of observing multiple continuous variables simultaneously. It is used to study the relationships between different variables and how they are distributed. Unlike a univariate distribution, which describes the probability of a single variable, a multivariate distribution takes into account the interactions between multiple variables.

2. What are some examples of continuous multivariate distributions?

Examples of continuous multivariate distributions include the multivariate normal distribution, multivariate t-distribution, and multivariate exponential distribution. These distributions are commonly used in fields such as economics, finance, and engineering to model complex systems with multiple variables.

3. How is a continuous multivariate distribution different from a discrete multivariate distribution?

A continuous multivariate distribution describes the probability of observing continuous variables, while a discrete multivariate distribution describes the probability of observing discrete variables. This means that the variables in a continuous multivariate distribution can take on any value within a certain range, while the variables in a discrete multivariate distribution can only take on specific values.

4. What is the importance of understanding continuous multivariate distributions?

Understanding continuous multivariate distributions is crucial in many fields of science, as it allows for the analysis and modeling of complex systems with multiple variables. It also provides insights into the relationships between different variables and how they affect each other. This knowledge can be used to make predictions, identify patterns, and make informed decisions.

5. How are continuous multivariate distributions used in data analysis?

Continuous multivariate distributions are used in data analysis to model and analyze data sets with multiple variables. They are often used in statistical tests and machine learning algorithms to identify patterns and relationships between variables. They can also be used to generate simulations and predictions, providing valuable insights for decision making.

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