Most probable value given observation

[Karnage1993]

Suppose I have observed $Z = 3$, where $Z = X + Y$, where $X \sim N(0,9), Y \sim N(0,4)$. How would I find the most probable value of $X$ that would have given me $Z = 3$?

My attempt at a solution: I was given that $X$ and $Y$ are independent, so that means $Z \sim N(0+0, 9+4) = N(0,13)$. To find the most probable value of $X$, we would have to find the highest possible probability of $Z$. But $Z$ is not discrete, so every probability at each point is 0, which means the highest probability of $Z$ is 0. Not sure what I would do after this, so I took a different direction.

Now I'm trying to find $E(X|Z=3)$ because I would think the mean is the best measure for the "most probable value of X". We have,

$E(X|Z=3) = E(X|X+Y=3)$. At this point, are there properties for conditional expectation involving independent random variables to answer this? I tried to find some but wasn't able to. Any help is appreciated.

 

[mfb]

But Z is not discrete, so every probability at each point is 0

Yes, but you can still look at the probability density function.

The expectation value does not have to be the most probable value.

 

[RUber]

My first instinct is to say that the most likely value of y is 0, so x, having larger standard deviation would be more likely to be 3. However, $p(x=3 \cap y=0) =.033$. Slightly higher values of y will likely optimize this function.
in general p(xy)=p(x)*p(y). The probability of a discrete value is not zero.

 

[FactChecker]

Slightly higher values of y will likely optimize this function.

Good catch!

The probability of a discrete value is not zero.

I think it's better to say that the probability density function is not zero. The probability of any single exact number is zero.

 

[WWGD]

To the OP: I think you want to look at the area of maximum likelihood estimation

http://en.wikipedia.org/wiki/Maximum_likelihood

 

[Ray Vickson]

Suppose I have observed $Z = 3$, where $Z = X + Y$, where $X \sim N(0,9), Y \sim N(0,4)$. How would I find the most probable value of $X$ that would have given me $Z = 3$?

My attempt at a solution: I was given that $X$ and $Y$ are independent, so that means $Z \sim N(0+0, 9+4) = N(0,13)$. To find the most probable value of $X$, we would have to find the highest possible probability of $Z$. But $Z$ is not discrete, so every probability at each point is 0, which means the highest probability of $Z$ is 0. Not sure what I would do after this, so I took a different direction.

Now I'm trying to find $E(X|Z=3)$ because I would think the mean is the best measure for the "most probable value of X". We have,

$E(X|Z=3) = E(X|X+Y=3)$. At this point, are there properties for conditional expectation involving independent random variables to answer this? I tried to find some but wasn't able to. Any help is appreciated.

Look at
$$f(x|z) \equiv \lim_{\Delta x \to 0, \Delta z \to 0} P(x < X < x + \Delta x\,| z < Z < z + \Delta z)\\ = \lim_{\Delta x \to 0, \Delta z \to 0} \frac{P(x < X < x + \Delta x \: \cap \: z < Z < z + \Delta z)}{P(z < Z < z + \Delta z)}$$
For $z = 3$ this will give you
$$c \, f_X(x) f_Y(3-x)$$
where $c$ is a normalization constant. For what value of $x$ would that be maximized?