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Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $Y \sim N(1,2)$ then what is the distribution of $2Y$? Well, we double the mean and multiply the variance by $2^2$, so $Y \sim N(2,8)$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $Z = X + (-Y)$ instead?

(In fact this is basically the same argument as pointed out in an older question as pointed out by Dilip Sarwate: https://stats.stackexchange.com/a/31328/6633)

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $Y \sim N(1,2)$ then what is the distribution of $2Y$? Well, we double the mean and multiply the variance by $2^2$, so $Y \sim N(2,8)$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $Z = X + (-Y)$ instead?

(In fact this is basically the same argument as pointed out in an older question by Dilip Sarwate: https://stats.stackexchange.com/a/31328/6633)

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $Y \sim N(1,2)$ then what is the distribution of $2Y$? Well, we double the mean and multiply the variance by $2^2$, so $Y \sim N(2,8)$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $Z = X + (-Y)$ instead?

(In fact this is basically the same argument as pointed out in an older question as pointed out by Dilip Sarwate: http://stats.stackexchange.com/a/31328/6633)

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $Y \sim N(1,2)$ then what is the distribution of $2Y$? Well, we double the mean and multiply the variance by $2^2$, so $Y \sim N(2,8)$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $Z = X + (-Y)$ instead?

(In fact this is basically the same argument as pointed out in an older question as pointed out by Dilip Sarwate: https://stats.stackexchange.com/a/31328/6633)

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Ok, since this is homework, you get hints instead if straight answers.

Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you just need to work out the distribution of $Z=X-Y$

Do you know how to do that?

Edit

Ok, so your problem is with the distribution of the difference. Try this:

If $Y \sim N(1,2)$ then what is the distribution of $2Y$? Well, we double the mean and multiply the variance by $2^2$, so $Y \sim N(2,8)$. Notice that this ensures that the spread of the distribution (standard deviation) has doubled, which makes sense. Now you know how to add random variable so what happens if you do $Z = X + (-Y)$ instead?

(In fact this is basically the same argument as pointed out in an older question as pointed out by Dilip Sarwate: http://stats.stackexchange.com/a/31328/6633)