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Normal Distributions |
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| Nov4-08, 02:35 PM | #1 |
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Normal Distributions
Hi say I have two "independent" Normal distributions,
S ~ N(0,3^2) and D~(0,2^2) since I know that S and D are indpendent then P(S ) + P(D) = P(S)P(D) however we know they are both normal distributed so I amm just wondering what the general rule is for multiplying two normal distributions thanks |
| Nov4-08, 02:47 PM | #2 |
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I'm not sure what you mean by
[tex] P(S) + P(D) = P(S) P(D) [/tex] Are you trying to say that when normal random variables are added, the resulting random variable is their product? Not true. If [tex] \begin{align*} S & \sim n(\mu_S, \sigma^2_S)\\ D & \sim n(\mu_D, \sigma^2_D) \end{align*} [/tex] and they are independent, then the sum [tex] S + D [/tex] is normal, with mean [tex] \mu_S + \mu_D [/tex] and variance [tex] \sigma^2_S + \sigma^2_D [/tex] A similar result is true even if the two variables have non-zero correlation (the formula for the variance of the sum involves the correlation). If by 'product' [tex] P(S) P(D) [/tex] you mean the convolution of the distributions, you could go through that work, but it leads you to the same result I quoted above. |
| Nov5-08, 09:14 AM | #3 |
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P(S & D) = P(S)P(D) where "S" here really means a statement along the lines of "S lies between A and B", and similarly for "D". |
| Nov5-08, 09:47 AM | #4 |
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Normal Distributions
For events A and B, normally distributed or not, P(A&B)= P(A)P(B|A)= P(B)P(B|A) where P(A|B) and P(B|A) are the "conditional probabilities" : P(A|B) is "the probability that A will happen given that B happened" and P(B|A) is "the probability that B will happen given that A happened".
IF the A and B are independent then P(A|B)= P(A) and P(B|A)= P(B) so you just multiply the separate probabilities. If they are not independent, just knowing the probabilities of each separately is not enough. You must know at least one of P(A|B), P(B|A) or P(A&B) separately from the individual probabilities. |
| Nov5-08, 10:59 AM | #5 |
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I answered as I did because
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