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Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}\begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1] \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}\begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2] \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [4] \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$$\int_{0}^1 c(u,b) db$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1. 

  • I don't think that $g(u)$ equals 1?
  • Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.
  • With respect to copula's, I'm not sure what the interpretation of $\int_{0}^1 c(u,b) db$ even means? Because the copula captures dependency between random variables, I don't know if looking for meaning there is fruitful?

Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.

Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1] \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [2] \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [3] \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [4] \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b) db$ is the marginal of this density, lets call it $g(u)$.  

  • I don't think that $g(u)$ equals 1?
  • Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.
  • With respect to copula's, I'm not sure what the interpretation of $\int_{0}^1 c(u,b) db$ even means? Because the copula captures dependency between random variables, I don't know if looking for meaning there is fruitful?
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Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.

Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly?

Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly? Looking at the probabilistic perspective, if $v=1$, then we have $P[V \leq 1 | U = u]$, which I think always evaluates to 1.

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Partial Derivative of Joint Distribution Function interpretation

Suppose we have \begin{equation} F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \end{equation}

From this, we can say the following: \begin{align} \frac{\partial F(x,y)}{\partial x} &= \frac{\partial}{\partial x} \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \\ & = \int_{-\infty}^y f(x,b) \ db \end{align}

The interpetation of this is nice, if $y = \infty$ (assuming $x,y \in [-\infty, \infty]$), then $\frac{\partial F(x,y)}{\partial x} = f(x)$. This can be seen by both the derivative form of $F(x,y)$ and the integral form of $f(x,y)$.

1.) Is it correct to say that the probabilistic interpretation of this is $P[Y \leq y | X = x]$? (I got this and adapted it from Nelsen's Inntroduction to Copula's book).

Now, we also know that a bivariate Copula function is also a joint distribution function. To repeat, let us have

\begin{equation} C(u,v) = \int_{0}^u \int_{0}^v c(a,b) \ db \ da \end{equation}

\begin{align} \frac{\partial C(u,v)}{\partial u} &= \frac{\partial}{\partial u} \int_{0}^u \int_{0}^v c(a,b) \ db \ da \\ & = \int_{0}^v c(u,b) \ db \\ & = P[V \leq v | U = u] \end{align}

2.) If $v = 1$, then something peculiar seems to happen. We know from the properties of copulas that $C(u,1) = u$, which would mean that $\frac{\partial C(u,v)}{\partial u}\vert_{v=1} = \frac{\partial C(u,1)}{\partial u} = \frac{\partial u}{\partial u} = 1$. However, looking at it from the integral form we have $\int_{0}^v c(u,b) \ db \vert_{v=1} = \int_{0}^1 c(u,b) \ db $. Now, technically, because $c(u,v)$ is a valid joint density function, $\int_{0}^1 c(u,b)$ is the marginal of this density, lets call it $g(u)$. I don't think that $g(u)$ equals 1? Or am I performing the partial derivative incorrectly?