# Simple problem

• Oct 16th 2010, 02:18 PM
financial
Simple problem
I'm stuck but here's what I have so far. I'm trying to find the probability that S is bigger than T.

$T \sim N(18,4), S \sim N(20,1)$

$Pr \{S>T \} = Pr \{ \frac{S-20}{1} > \frac{T-20}{1} \}$

$=Pr\{N(0,1)>T-20\}$

$=Pr\{T-20

$\mbox{Let }X=T-20 \mbox{ then } X \sim N(18-20,4) \sim N(-2,4)$

$\Rightarrow Pr\{T-20

$=Pr\{X

$=Pr\{N(-2,4)

Where do I go from here?
• Oct 16th 2010, 02:26 PM
mr fantastic
Quote:

Originally Posted by financial
I'm stuck but here's what I have so far. I'm trying to find the probability that S is bigger than T.

$T \sim N(18,4), S \sim N(20,1)$

$Pr \{S>T \} = Pr \{ \frac{S-20}{1} > \frac{T-20}{1} \}$

$=Pr\{N(0,1)>T-20\}$

$=Pr\{T-20

$\mbox{Let }X=T-20 \mbox{ then } X \sim N(18-20,4) \sim N(-2,4)$

$\Rightarrow Pr\{T-20

$=Pr\{X

$=Pr\{N(-2,4)

Where do I go from here?

Are S and T independent? I suppose so. Then you should consider the random variable W = S - T and calculate Pr(W > 0).

Quote:

Originally Posted by Wikipedia at en.wikipedia.org/wiki/Normal_distribution
if X1, X2 are two independent normal random variables, with means μ1, μ2 and standard deviations σ1, σ2, then their linear combination will also be normally distributed:

• Oct 17th 2010, 03:13 PM
financial
Wow, why couldn't I think of that myself?

So now I have:

$\mbox{Let }X=S-T$

$X \sim N(20-18, 1+4) \sim N(2, 5)$

$Pr\{S>T\} = Pr\{W>0\}= Pr\{ \frac{W-2}{\sqrt{5}}>\frac{0-2}{\sqrt{5}} \}$

$=Pr\{N(0,1) > \frac{-2}{\sqrt{5}} \}=\Phi(\frac{-2}{\sqrt{5}})$

Is this correct?
• Oct 17th 2010, 04:47 PM
financial
On a side note, what if they're correlated?

Let's say the covariance between S and T is 1.2. Now, how do I calculate the probability that S is greater than T?
• Oct 17th 2010, 04:56 PM
mr fantastic
Quote:

Originally Posted by financial
On a side note, what if they're correlated?

Let's say the covariance between S and T is 1.2. Now, how do I calculate the probability that S is greater than T?

Then obviously you need to review the difference of two correlated normal distributions: http://www.stanford.edu/~srabbani/bivariate.pdf
• Oct 17th 2010, 05:04 PM
financial
Okay this time I get

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And continue the same calculation as previously.