# Can a random variable donimate

1. Jul 22, 2010

### jillna

Given two random variables x and y, and a constant c

What conditions are needed to make:

$$Prob( w x + y < c ) \approx Prob( w x < c ), \text{ for } w \rightarrow \infty$$

Can anyone help? I think $$E(x) < \infty$$ and $$E(y) < \infty$$ might do. Is this right?

tks!!!!

Last edited: Jul 22, 2010
2. Aug 3, 2010

### Eynstone

I think you need the expected value of y an order of magnitude less than E(x).

3. Aug 4, 2010

### brian44

In the limit as $$w \rightarrow \infty$$ I believe they are always equal. I will use the probability density functions (f(x),f(y), and f(x,y)) to give my reasoning.

$$P(wx < c) = P(x < c/w) = P(x < 0)$$ in the limit of $$w \rightarrow \infty$$
$$= \int_{-\infty}^{0}f(x)dx = \int_{-\infty}^{0}\int_{-\infty}^{\infty}f(x,y)dydx$$

To calculate the probability you have to add up the region of the density for which wx+y < c, which can be achieved by integrating for each x from y=-infinity to the line y=-wx+c:

$$P(wx +y < c) = \int_{-\infty}^{\infty}\int_{-\infty}^{-wx+c}f(x,y)dydx$$

In the limit this becomes the y-axis, so we actually have in this case:
$$= \int_{-\infty}^{0}\int_{-\infty}^{\infty}f(x,y)dydx = \int_{-\infty}^{0}f(x)dx$$

and so they are equivalent.

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