3 added 390 characters in body

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

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

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

\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 $$$$F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da$$$$

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

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

\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 $$$$F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [1]$$$$

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

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

\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?

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

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

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

\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 $$$$F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da$$$$

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

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

\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 $$$$F(x,y) = \int_{-\infty}^x \int_{-\infty}^y f(a,b) \ db \ da$$$$

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

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

\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.

1

# Partial Derivative of Joint Distribution Function interpretation

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

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

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

\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?