$(X|Y=y)\sim N(y,y^2)$, $Y\sim U[3,9]$. Find $Var(X).$

Question

$(X|Y=y)\sim N(y,y^2)$, $Y\sim U[3,9]$, where $N(y,y^2)$ is a normal distribution with mean $y$ and variance $y^2$, $U[3,9]$ is a uniform distribution on $[3,9].$

On this condition, find $Var(X).$

My attempt:

Use the law of total variance : $Var(X) = E(Var(X|Y))+Var(E(X|Y)).$

Since $(X|Y=y)\sim N(y,y^2)$, $E(X|Y)=Y$ and $Var(X|Y)=Y^2$ (?)

Plug them into the law : $Var(X)=E(Y^2)+Var(Y).$

Since $Y\sim U[3,9]\implies Var(Y)=\frac{(9-3)^2}{12}=3, E(Y)=\frac{9+3}{2}=6\implies E(Y^2)=Var(Y)+E(Y)^2=3+6^2=39.$

$\implies Var(X)=39+3=42.$

The thing is, I'm not sure about the part I marked with (?).

I'm comfortable with saying that $(X|Y=y)\sim N(y,y^2)\implies E(X|Y=y)=y,\ Var(X|Y=y)=y^2$ because it's just what it is.

But I don't quite feel right about saying that $(X|Y=y)\sim N(y,y^2)\implies E(X|Y)=Y,\ Var(X|Y)=Y^2$, because $y$ is just a number and $Y$ is a random variable. (and of course $E(X|Y=y)$ and $E(X|Y)$ are different things, I guess)

Actually my answer is correct, and it makes me feel more uncomfortable since I don't know what exactly I'm doing right now.

More specifically, I know by the definition that $E(X|Y=y)=\int x\frac{f(x,y)}{f_Y(y)}dx$, but not sure what $E(X|Y)$ means. Is $E(X|Y)=\int x\frac{f(x,Y)}{f_Y(Y)}dx$ a valid expression? But does $f(x,Y)$ make sense? And what's the relationship between $(X|Y=y)$ and $X|Y$? The right side of $|$ symbol only affects to the PDF?

I want to know what I'm doing. Any help would be appreciated.

Answer 1

Your calculation and reasoning are correct.

The conditional distribution of $X$ given $Y$ is normal with mean $Y$ and variance $Y^2$; so by construction, what this means is $$\operatorname{Var}[X \mid Y] = Y^2.$$ In fact, the property that $X \mid Y$ is normally distributed is not relevant to the total variance calculation. So long as $X \mid Y$ is a random variable whose mean is $Y$ and variance $Y^2$, you will get the same result for $\operatorname{Var}[X]$.

I am a bit puzzled as to why you accept $\operatorname{E}[X \mid Y = y] = y$, yet not $\operatorname{E}[X \mid Y] = Y$. The latter has simply replaced a deterministic variable with a random one. The result is that the moment is itself a random variable. So when we write something like $\operatorname{Var}[X \mid Y]$, we are talking about a random variable that is a function of the random variable $Y$. For example, $W = Y^2$ is also a function of the random variable $Y$. I could even write something like $$\int_{x=0}^\infty Y x^2 e^{-Y x} \, dx$$ and this is a random variable that is a function of $Y$. In fact, it is $$\operatorname{E}[X^2 \mid Y]$$ when $X \mid Y \sim \operatorname{Exponential}(Y)$ where $Y$ is a rate parameter. So long as $Y \ge 0$ this integral and the resulting conditional expectation, is well-defined.