# Conditional probability for normally R.V

1. Jan 14, 2014

### ehudwe

w1,w2,w3,w4,w5 are normally distributed i.i.d R.V.

I want to calculate the probability that slots 1&2(w1,w2) are smaller than slots 3,4,5 (where mu>0)

I'm able to calculate the probability that slot 1 is smaller than 3,4,5, now i'm stuck on the calculation for slot 2, where i know that i have to consider the outcomes from the first calculation, for example if i know that w1(slot 1) is larger than w2(slot 2), than the probability that w2(slot 2) is smaller than (3,4,5) is 1.

until now i was sure that the whole probability is :

$$P_{correct}= P(w_1<\mu+w_3)P(w_1<\mu+w_4)P(w_1<\mu+w_5)P(w_2<\mu+w_3)P(w_2<\mu+w_4)P(w_2<\mu+w_5)$$
now i am thinking of:
$$P_{correct}= P(w_1<\mu+w_3)P(w_1<\mu+w_4)P(w_1<\mu+w_5)P(w_2<\mu+w_3|w_1<w_2)P(w_2<\mu+w_4|w_1<w_2)P(w_2<\mu+w_5|w_1<w_2)P(w_2<\mu+w_3|w_2<w_1)P(w_2<\mu+w_4|w_2<w_1)P(w_2<\mu+w_5|w_2<w_1)$$where the last 3 terms are equal to 1 because i have the knowledge that $$P(w1<μ+w3)P(w1<μ+w4)P(w1<μ+w5)$$
any help will be appreciated.

2. Jan 15, 2014

### ehudwe

How do i edit my post ? i want to revise the question

3. Jan 16, 2014

### ehudwe

rewrite my question, i didn't find how to just edit

the following modulation is in the frequency domain, x-axis is frequency , y-axis is amplitude of electric current.

the signal is a pulse divided to 5 equally slots by means of electrical current where 2 "holes" indicating the binary message.
for this example my constellation size is 2^3. therefore 5C2 = 10 combinations is possible in this setup (5 slots and 2 "holes") and i'll will use 8 of them.

i am trying to calculate probability as described :

w1,w2,w3,w4,w5 are normally distributed i.i.d R.V (AWGN).

A correct detection is when the noise in slots 1 and 2 both is smaller than the noise + mu(the current related to the slot).

$$w_1,w_2<min(\mu+w_3,\mu+w_4,\mu+w_5)$$

so the probability for correct detection as i'm thinking is:

$$1. P_{correct}= P(w_1<\mu+w_3\cap w_1<\mu+w_4\cap w_1<\mu+w_5\cap w_2<\mu+w_3\cap w_2<\mu+w_4\cap w_2<\mu+w_5)$$

if i'm right, i can calculate just one of the terms , because of the i.i.d properties.
$$2. P_{correct}= P(w_1<\mu+w_3)P(w_1<\mu+w_4)P(w_1<\mu+w_5)P(w_2<\mu+w_3)P(w_2<\mu+w_4)P( w_2<\mu+w_5)$$
$$P_{correct}= P(w_1<\mu+w_3)^6$$
where
$$P(w_1<\mu+w_3)= \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{w_1^2}{2\sigma^2}}Q(\frac{w_1-\mu}{\sigma})dw_1$$
hence the error probability is :
$$P_{error}=1-P_{correct}$$
therefore the general case with n- slots and k- "holes" the probability is
$$3.P_{correct}=\left[\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}Q(\frac{x-\mu}{\sigma})dx\right]^{(n-k)k}$$
i did another calculation with my friend and we got:
$$4.P_{correct}=\left[\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{x^2}{2\sigma^2}}\left[Q(\frac{x-\mu}{\sigma})\right]^{(n-k)}dx\right]^k$$

i'm trying to understand which of Eq.3 or Eq.4 is the correct answer ? and the transition from Eq.1 to Eq.2 is legal, meaning, are slots 1,2,3,4,5 are absolutely independent ?

4. Jan 16, 2014

### Stephen Tashi

A way to put that in familiar territory is:

$max(w_1,w_2) < min(\mu+w_2,\mu+w_4,\mu+w_5)$

Look at the theory of "order statistics" to find out how the max and min are distributed.

5. Jan 19, 2014

### ehudwe

thanks i'm hoping it will do the work.