0
$\begingroup$

Assume we have two continuous variables $X,Y$; and we know the conditional probability density function: $f_{x|y}(x,y)$

How might we calculate the conditional probability over a range of values: $P(x_0 \leq x \leq x_1|y_0 \leq y \leq y_1)$

I know we could use the definition: $P(x_0 \leq x \leq x_1|y_0 \leq y \leq y_1) = \frac{ \int_{x_0}^{x_1}\int_{y_0}^{y_1}f(x,y)dydx }{ \int_{y_0}^{y_1} f(y)}$ where $f(y)$ and $f(x,y)$ are the probability distribution functions of the just $Y$ and then $X,Y$, the joint.

I can calculate $f(y)$ quickly, but I would rather not calculate $f(x,y)$. This is for a numerical application and calculating $f(x,y)$ is expensive. Is there a way to use the the conditional probability density function directly? I feel like the answer is a simple integral; but this is my first attempt at dealing with continuous and discrete probability density functions in the context of conditional/joint probabilities.

If someone has a good reference that helped one "graduate" from the low-level undergraduate style probability (nearly everything is discrete events) to working with continuous or discrete density functions directly, I would be most appreciative.

$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

If you know $f_{x|y}(x,y)$ and $f(y)$ then

Define $h(x) = \frac{\int_{y_0}^{y_1}{f_{x|y}(x,y)f(y)}dy}{\int_{y_0}^{y_1}{f(y)}dy}$

$P[x_0 \leq X \leq x_1 | y_0 \leq Y \leq y_1] = \int_{x_0}^{x_1}{h(x)dx}$

$\endgroup$
5
  • $\begingroup$ Thank-you! Is there a theorem that gives the right hand side? It is not clear to me why we can define $h(x)$ in that way. $\endgroup$
    – Vance T
    Apr 26, 2020 at 21:23
  • $\begingroup$ It looks like $h(x)$ is using the "definition" of probability? The numerator is another way of stating the joint probability? $\endgroup$
    – Vance T
    Apr 26, 2020 at 21:32
  • $\begingroup$ @VanceT that’s right.... I used this approach because you said that the joint probability is computationally expensive to derive. $\endgroup$
    – Vishaal Sudarsan
    Apr 26, 2020 at 21:47
  • $\begingroup$ @VanceT h(x) is basically the pdf of x given y belongs to that range. $\endgroup$
    – Vishaal Sudarsan
    Apr 26, 2020 at 21:49
  • $\begingroup$ Thank-you for your help and clarifying comments! $\endgroup$
    – Vance T
    Apr 26, 2020 at 21:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .