Statistics: Probability of False Negative during Measurement

[n00bcake22]

Hello Everyone,

My statistics is terribly rusty so I am turning to all of you for assistance! I am in the process of reviewing my old text but I figured this may be quite a bit quicker.

Homework Statement

Suppose "x" is normally distributed with "mu_1" and "sigma_1." Now suppose x is measured with a device whose output is also normally distributed where "mu_2" equals the true value of x and has a standard deviation of "sigma_2."

I am trying to figure out how to find the total probability that the measurement device will say x < "a" (some value < mu_1) when in fact x >= a (i.e. a false negative).

If that makes sense...

Homework Equations

The Attempt at a Solution

I know how to determine P(x < a) for the x-distribution alone and could determine the probability of the false negative if I was given a particular, known x-value but I have no idea how to find the TOTAL probability of false negatives when x >=a but not exactly known. This has been driving me crazy all morning.

Thanks in advance Everyone!

 

[lanedance]

for a given x value the probabilty of a false negative will be the "tail" of the measurement distribution that spreads below a.

So call a mearuement y
y = x+e
where
x is the actual quatinty to be measured (Normallly distributed RV, N(mu_1, sigma_1))
x is the actual quatinty to be measured (Normallly distributed RV, (0, sigma_2)))

so as discussed you should be able to find
P(y<a|x)

now sum this over all possible x and its probabilty distribution

 

[Stephen Tashi]

I think the probability of the event { 'a' is greater than or equal to mu1 and the reading is less than or equal to mu1 } is:

\int_{\mu_1}^\infty \frac{1}{\sqrt{2\pi} \sigma_1 } e^\frac{-(a-\mu_1)^2}{2\sigma_1^2} <br /> \big{[} \int_{-\infty}^{\mu_1 - a} \frac{1}{\sqrt{2\pi}\sigma_2}e^\frac{-(x-\mu_2)^2}{2\sigma_2^2} dx \big{]} da

I don't recognize this expression as anything you could look up in standard statistical tables. It can be computed numerically. I'd bet there are papers written about this type of problem. We just have to find the right keywords for a search.

 

[n00bcake22]

@Stephen: I think you read my question wrong as your description doesn't match my statement. Lanedance has the correct idea.

x = population = N(mu1, sig1)
y = measurement of x = x_true + e
e = error = N(0, sig2)
a = some lower bound, a < mu1

I would like to know the total probability of false negatives provided that the true value of x >= a (i.e. what is the total probability that for any x >= a, y < a). I think it would look something like this in statistical syntax (wild guess)...

P((y < a)|(x >= a))

So I can calculate P(y<a) at x = a, x = a + dx, x = a + 2*dx, ..., and sum them all up but this doesn't seem right. How do I incorporate the PDF of x itself?

 

[Stephen Tashi]

I'm not going to call the measurement error 'e' because of the confusion with the number 'e'. I'll call the measurement error 'w'.

Let \sigma_3 = \sqrt{ \sigma_1^2 + \sigma_2^2}

Let \mu_3 =\mu_1

Let C = \frac{1}{ \sqrt{2\pi} \sigma_3} \int_a^\infty e^ \frac{-(y-\mu_3)^2}{2 \sigma_3^2} dy

p(x \leq a | y \geq a) = p(x \leq a |x + w \geq a) = p( x \leq a and x + w \geq a)/ p(x+w \geq a) =

\frac{1}{C} \int_{-\infty}^a \frac{1}{\sqrt{2\pi} \sigma_1 } e^\frac{-(x-\mu_1)^2}{2\sigma_1^2} <br /> \big{[} \int_{a-x}^{\infty} \frac{1}{\sqrt{2\pi}\sigma_2}e^\frac{-w^2}{2\sigma_2^2} dw \big{]} dx

 

[Stephen Tashi]

I see that I answered the wrong question, in my last post.Let C = \frac{1}{ \sqrt{2\pi} \sigma_1} \int_a^\infty e^ \frac{-(x-\mu_1)^2}{2 \sigma_1^2} dy

What you asked was:
p(y &lt; a | x \geq a) = p(x + w &lt; a | x \geq a) = p( x + w &lt; a and x \geq a)/ p(x \geq a) =

\frac{1}{C} \int_a^\infty \frac{1}{\sqrt{2\pi} \sigma_1 } e^\frac{-(x-\mu_1)^2}{2\sigma_1^2} <br /> \big{[} \int_{-\infty}^{x-a} \frac{1}{\sqrt{2\pi}\sigma_2}e^\frac{-w^2}{2\sigma_2^2} dw \big{]} dx