MHB Prove that lim y->infinity F(X,Y) (x,y) = F(X)(x)

[oyth94]

Prove that lim y->infinity F(X,Y) (x,y) = F(X)(x)

 

[TheBigBadBen]

re: Prove that lim y->infinity F(X,Y) (x,y) = F(X)(x)

Prove that lim y->infinity F(X,Y) (x,y) = F(X)(x)

What is F(X,Y) here? Is this the cumulative distribution function?

Are we given any information about X and Y in the problem?

 

[oyth94]

What is F(X,Y) here? Is this the cumulative distribution function?

Are we given any information about X and Y in the problem?

I think this is the joint cdf, This question is related to joint probability in some way.

i did lim y->infinity F_X,Y(x,y) = P(X<= x, Y <= infinity)
after that i am not sure if i skipped a step or went the wrong direction but the next step i did was conclude that
= P(X<=x)
= F_(X)(x)

please help!

 

[chisigma]

Prove that lim y->infinity F(X,Y) (x,y) = F(X)(x)

By definition is...

$$F_{X,Y} (x,y) = P \{X<x,Y<y\}$$

... and because if y tends to infinity then $P \{Y<y\}$ tends to 1 we have...

$$\lim_{y \rightarrow \infty} F_{X,Y} (x,y) = P \{X<x\} = F_{X} (x)$$

Kind regards

$\chi$ $\sigma$

 

[oyth94]

By definition is...

$$F_{X,Y} (x,y) = P \{X<x,Y<y\}$$

... and because if y tends to infinity then $P \{Y<y\}$ tends to 1 we have...

$$\lim_{y \rightarrow \infty} F_{X,Y} (x,y) = P \{X<x\} = F_{X} (x)$$

Kind regards

$\chi$ $\sigma$

So I did it correctly? No steps skipped?

 

[TheBigBadBen]

So I did it correctly? No steps skipped?

I would think that unless your professor wants you to prove this using $\epsilon$s and $\delta$s, you've said as much as you need to say. However, the best way to know is to ask the person who will be grading your homework.

 

[oyth94]

There is a similar question. But this one is not a limit question but does invoke joint probability, independence etc
The question is:

F_(X,Y(x,y) <= F_X(x),F_Y(y)

I know that when finding the integral for F_X(x) it is in respect to y. And when finding the integral for F_Y(y) we find integral with respect to x. When we multiply the two together to find if it is independent we multiply the two together and see if it equals to F_X,Y(x,y)
But I'm not sure if this question is regarding independence or something else. How must I go about proving this question?

 

[TheBigBadBen]

There is a similar question. But this one is not a limit question but does invoke joint probability, independence etc
The question is:

F_(X,Y(x,y) <= F_X(x),F_Y(y)

I know that when finding the integral for F_X(x) it is in respect to y. And when finding the integral for F_Y(y) we find integral with respect to x. When we multiply the two together to find if it is independent we multiply the two together and see if it equals to F_X,Y(x,y)
But I'm not sure if this question is regarding independence or something else. How must I go about proving this question?

I don't think that the premise of the question is true in general, you would have provide more information about the distribution on $X$ and $Y$.

 

[oyth94]

I don't think that the premise of the question is true in general, you would have provide more information about the distribution on $X$ and $Y$.

This was all that was given in the question. So I am confused now... Or can we prove by contradiction if possible?

 

[TheBigBadBen]

This was all that was given in the question. So I am confused now... Or can we prove by contradiction if possible?

Sorry, I was a little hasty with that; let me actually be sure about what the question states. My understanding is that you are to prove that for any $X,Y$ with some joint cdf $F_{X,Y}$ and where $X$ and $Y$ are not necessarily independent, we can state that
$$
F_{X,Y}(x,y)\leq F_X(x)\cdot F_Y(y)
$$
Or, phrased in a different way:
$$
P(X< x \text{ and } Y< y) \leq P(X < x)\cdot P(Y < y)
$$

If the above is what you meant, I would pose the following counterargument: we could equivalently state
$$
P(X< x|Y<y) \cdot P(Y< y) \leq P(X < x)\cdot P(Y < y) \Rightarrow \\
P(X<x|Y<y) \leq P(X<x)
$$
and that simply isn't true for all distributions. That is, prior knowledge of another variable can increase the probability of an event. Would you like a counter-example to this claim?

If that's not what you meant, or if there's more information about $X$ and $Y$ that you're leaving out, do say so.

 

[TheBigBadBen]

I realize what you conceivably could have meant (and probably did mean) is

$$
\text{for any }y:\; F_{X,Y}(x,y)\leq F_X(x) \text{ AND }\\
\text{for any }x:\; F_{X,Y}(x,y)\leq F_Y(y)
$$

Is this what you meant to prove? Then yes, we can prove this by integration, as you rightly mentioned.

Please, please, please: try to be clearer in the future about what you mean, even if it makes your post a little longer.

 

[oyth94]

I realize what you conceivably could have meant (and probably did mean) is

$$
\text{for any }y:\; F_{X,Y}(x,y)\leq F_X(x) \text{ AND }\\
\text{for any }x:\; F_{X,Y}(x,y)\leq F_Y(y)
$$

Is this what you meant to prove? Then yes, we can prove this by integration, as you rightly mentioned.

Please, please, please: try to be clearer in the future about what you mean, even if it makes your post a little longer.

Hi my apologies this is actually what I meant to say. So how does it work after integration? An I doing: the Integral from 0 to y for FX(x)dy multiply with integral from 0 to x of FY(y)dx to get integral of FXY(x,y)? Okay something is wrong I don't think that makes sense does it?

 

[oyth94]

Re: Prove that lim y-&gt;infinity F(X,Y) (x,y) = F(X)(x)

Sorry, I was a little hasty with that; let me actually be sure about what the question states. My understanding is that you are to prove that for any $X,Y$ with some joint cdf $F_{X,Y}$ and where $X$ and $Y$ are not necessarily independent, we can state that
$$
F_{X,Y}(x,y)\leq F_X(x)\cdot F_Y(y)
$$
Or, phrased in a different way:
$$
P(X< x \text{ and } Y< y) \leq P(X < x)\cdot P(Y < y)
$$

If the above is what you meant, I would pose the following counterargument: we could equivalently state
$$
P(X< x|Y<y) \cdot P(Y< y) \leq P(X < x)\cdot P(Y < y) \Rightarrow \\
P(X<x|Y<y) \leq P(X<x)
$$
and that simply isn't true for all distributions. That is, prior knowledge of another variable can increase the probability of an event. Would you like a counter-example to this claim?

If that's not what you meant, or if there's more information about $X$ and $Y$ that you're leaving out, do say so.

For the counter argument why did you use conditional probability? And multiplied by p(Y<y)?

- - - Updated - - -

For the counter argument why did you use conditional probability? And multiplied by p(Y<y)?

Oh sorry never mind I understand why you used the conditional probability multiply with P(Y<y) because it is te intersection of X<x and Y<y. So that is the counter argument used against this proof?

 

[TheBigBadBen]

Hi my apologies this is actually what I meant to say. So how does it work after integration? An I doing: the Integral from 0 to y for FX(x)dy multiply with integral from 0 to x of FY(y)dx to get integral of FXY(x,y)? Okay something is wrong I don't think that makes sense does it?

So the proof of the first inequality via integrals would go something like this:
First of all, definition. We state that there is some (joint) probability density function of the form $f_{X,Y}(x,y)$. We can then supply the following definitions of $F_X,F_Y,$ and $F_{X,Y}$ in terms of integrals:
$$
F_{X,Y}(x,y)=\int_{-\infty}^y \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy\\
F_{X}(x)=\int_{-\infty}^{\infty} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy\\
F_{Y}(y)=\int_{-\infty}^{y} \int_{-\infty}^\infty f_{X,Y}(x,y)\,dx\,dy
$$

With that in mind, we may state that
$$
F_{X}(x)=\int_{-\infty}^{\infty} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy\\
= \int_{-\infty}^{y} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy +
\int_{y}^{\infty} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy
$$
Since $f_{X,Y}(x,y)$ is a probability distribution, it is non-negative at all $(x,y)$, which means that both of the above integrals are non-negative. This allows us to state that
$$
\int_{-\infty}^{y} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy +
\int_{y}^{\infty} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy \geq \\
\int_{-\infty}^{y} \int_{-\infty}^x f_{X,Y}(x,y)\,dx\,dy = F_{X,Y}(x,y)
$$
Thus, we may conclude that in general, $F_{X,Y}(x,y) \leq F_Y{y}$.

However, it is not necessary to appeal to these integral definitions in order to go through the proof. I'll post an alternate, simpler proof as well

 

[TheBigBadBen]

The Easier Proof (via set theory)

Consider the following definitions:
$$
F_{X,Y}(x,y)=P(X< x \text{ and } Y < y)\\
F_{X}(x)=P(X< x)\\
F_{Y}(y)=P(Y< y)
$$

We note that the set of $(x,y)$ such that $X< x \text{ and } Y < y$ is the intersection of the set of $(x,y)$ such that $X< x$ and the set of $(x,y)$ such that $Y < y$. Since this set is a subset of each set, the probability of the associated event is less than or equal to the probability of either of the other events.