# Condition Probability - What am I doing wrong?

I am reviewing some notes of mine and refreshing myself with some statistics and I came across a problem that asks me to calculate $$P(X>1|Y>1)$$ for the random variables $$X$$ and $$Y$$ whose joint probability density function is given by:

$$\begin{eqnarray*} f(x,y) & = & \begin{cases} ye^{-y(x+1)} & x>0,\,y>0\\ 0 & otherwise \end{cases} \end{eqnarray*}$$

I thought of two ways of computing this and I thought that regardless of which approach I took, I should end up with the same answer, but I'm getting two different final answers. I'm hoping someone could help me figure out which is the right answer AND why the other answer is not correct.

## Method 1:

$$\begin{eqnarray} P(X>1|Y>1) & = & \frac{P(X>1,\,Y>1)}{P(Y>1)}\nonumber \\ & = & \frac{\int_{1}^{\infty}\int_{1}^{\infty}f(x,y)dxdy}{\int_{1}^{\infty}f(y)dy} &&&&&&&&&&&(1) \end{eqnarray}$$

To calculate this, I first find the marginal pdf, $$f(y)$$ for the denominator in this last term:

$$\begin{eqnarray*} f(y) & = & \int_{0}^{\infty}ye^{-y(x+1)}dx\\ & = & ye^{-y}\int_{1}^{\infty}e^{-yx}dx\\ & = & \left(ye^{-y}\right)\left(\left.-\frac{1}{y}e^{-yx}\right|_{x=0}^{x=\infty}\right)\\ & = & \left(e^{-y}\right)\left(\left.-e^{-yx}\right|_{x=0}^{x=\infty}\right)\\ & = & \left(e^{-y}\right)\left(0+1\right)\\ & = & e^{-y} \end{eqnarray*}$$

so $$f(y)=e^{-y}$$ for $$y\in(0,\infty)$$. So using this, we now calculate $$P(Y>1)$$ (the denominator term in $$(1)$$ above):

$$\begin{eqnarray*} P(Y>1) & = &\int_{1}^{\infty}f(y)dy\\ & = &\int_{1}^{\infty}e^{-y}dy\\ & = & \left.-e^{-y}\right|_{y=1}^{y=\infty}\\ & = & 0+e^{-1}\\ & = & e^{-1} \end{eqnarray*}$$

Now, we compute the numerator in $$(1)$$:

$$\begin{eqnarray*} P(X>1,\,Y>1) & = & \int_{1}^{\infty}\int_{1}^{\infty}ye^{-y(x+1)}dxdy\\ & = & \int_{1}^{\infty}ye^{-y}\int_{1}^{\infty}e^{-xy}dxdy\\ & = & \int_{1}^{\infty}ye^{-y}\left(\left.-\frac{1}{y}e^{-yx}\right|_{x=1}^{x=\infty}\right)dy\\ & = & \int_{1}^{\infty}e^{-y}\left(0+e^{-y}\right)dy\\ & = & \int_{1}^{\infty}e^{-2y}dy\\ & = & \frac{1}{2}e^{-2} \end{eqnarray*}$$

Assembling the numerator and denominator components should give us the final answer:

$$\begin{eqnarray*} \frac{P(X>1,\,Y>1)}{P(Y>1)} & = & \frac{\frac{1}{2}e^{-2}}{e^{-1}} & = & \frac{1}{2}e^{-2}e^{1} & = & \frac{1}{2}e^{-1} \end{eqnarray*}$$

So, on to method 2:

## Method 2

This method differs from Method 1 in that I compute (I think) $$f(x|y)$$ and then compute the desired probability from the conditional pdf:

$$\begin{eqnarray*} f(x|y) & = & \frac{f(x,y)}{f(y)}\\ & = & \frac{ye^{-y(x+1)}}{e^{-y}}\\ & = & ye^{-y(x+1)}e^{y}\\ & = & \begin{cases} ye^{-xy} & x>0,\,y>0\\ 0 & otherwise \end{cases} \end{eqnarray*}$$

So, now, to calculate $$P(X>1|Y>1)$$, I compute the following:

$$\begin{eqnarray*} P(X>1|Y>1) & = & \int_{1}^{\infty}\int_{1}^{\infty}f(x|y)dxdy\\ & = & \int_{1}^{\infty}\int_{1}^{\infty}ye^{-xy}dxdy\\ & = & \int_{1}^{\infty}y\int_{1}^{\infty}e^{-xy}dxdy\\ & = & \int_{1}^{\infty}y\left(\left.-\frac{1}{y}e^{-xy}\right|_{x=1}^{x=\infty}\right)dy\\ & = & \int_{1}^{\infty}\left(\left.-e^{-xy}\right|_{x=1}^{x=\infty}\right)dy\\ & = & \int_{1}^{\infty}\left(e^{-y}\right)dy\\ & = & \left.-e^{-y}\right|_{y=1}^{y=\infty}\\ & = & 0+e^{-1}\\ & = & e^{-1} \end{eqnarray*}$$

So, with Method 1 I get $$P(X>1|Y>1)=\frac{1}{2}e^{-1}$$ and with Method 2 I get $$P(X>1|Y>1)=e^{-1}$$. So, which one is right AND, if I haven't just made a simple error somewhere, why don't both of these methods return the same result? Is there a fundamental flaw in Method 1 or Method 2?

Both methods give you the same answer, as they should.

In the first method, as you said,

\begin{align} P(X>1\mid Y>1)&=\frac{P(X>1,Y>1)}{P(Y>1)} \\&=\frac{1}{P(Y>1)}\iint\mathbf1_{x>1,y>1}f(x,y)\,dx\,dy \end{align}

In the second method,

\begin{align} P(X>1\mid Y>1)&=\int\mathbf1_{x>1}\color{blue}{f_{X\mid Y>1}(x)}\,dx \\&=\int\mathbf1_{x>1}\left(\color{blue}{\int\mathbf1_{y>1}\frac{ f(x,y)}{P(Y>1)}\,dy}\right)\,dx \end{align}

Your formulation of method 2 was not quite correct, hence the confusion.

note that $$P(X>1|Y>1)$$ is not equal $$\int_{1}^{\infty} \int_{1}^{\infty} f(x|y) dx dy$$ see modern probability for more information.

it is equal $$\int_{1}^{\infty} f(x|Y>1) dx$$ . you drive($$\frac{d}{dx}$$) from $$F(x|Y>1)$$ to calculate $$f(x|Y>1)$$.

so,

$$P(X>1|Y>1)=\int_{1}^{\infty} f(x|Y>1) dx =\int_{1}^{\infty} \frac{f(x,Y>1)}{f(Y>1)} dx= \frac{1}{f(Y>1)} \int_{1}^{\infty} f(x,Y>1)=\frac{1}{\int_{1}^{\infty} f(y)dy} \int_{1}^{\infty} \int_{1}^{\infty} f(x,y) dx dy=\frac{1}{\int_{1}^{\infty} f(y)dy} \int_{1}^{\infty} \int_{1}^{\infty} f(x|y)f(y) dx dy=\frac{1}{\int_{1}^{\infty} f_Y(t)dt} \int_{1}^{\infty} \int_{1}^{\infty} f(x|y)f(y) dx dy=\int_{1}^{\infty} \int_{1}^{\infty} f(x|y)\frac{f(y)}{\int_{1}^{\infty} f_Y(t)dt} dx dy \neq \int_{1}^{\infty} \int_{1}^{\infty} f(x|y) dx dy$$.

also

$$P(X>1|Y>1) = \frac{1}{\int_{1}^{\infty} f(y)dy} \int_{1}^{\infty} \int_{1}^{\infty} f(x,y) dx dy =\frac{1}{e^{-1}} \frac{1}{2}e^{-2}=\frac{1}{2}e^{-1}$$