Today’s Uncertainty Wednesday is a further continuation of the PSA Test example, but I got tired of the boring title and wanted to give the post as more, well, specific one. Last Wednesday I wrote that sensitivity and specificity are “widely used to assess the quality of medical tests.”

Here are their definitions taken from Wikipedia

Sensitivity = probability of a positive test given that the patient has the disease
Specificity = probability of a negative test given that the patient is well

Using our notation we can rewrite this as

Sensitivity = P(H | B), i.e. probability of a high PSA level signal (H) conditional on the patient having cancer (B)

Specificity = P(L | A), i.e. probability of a low PSA level signal (L) conditional on the patient being healthy (A)

We see that sensitivity is in a way the inverse of the question we were asking originally, which is P(B | H), i.e the probability of having cancer (B) conditional on a receiving a high PSA level signal (H).

So the question is what is the relationship between P(H | B) and P(B | H)?

To figure this out, let’s first see if we can derive P(H | B) from the elementary probabilities given originally, which I am repeating here

P({AL}) = P(healthy *and* low PSA) = 0.907179
P({AH}) = P(healthy *and* high PSA) = 0.089721
P({BL}) = P(cancer *and* low PSA) = 0.001519
P({BH}) = P(cancer *and* high PSA) = 0.001581

So let’t start by recalling out how likely it is that someone has cancer (B) which we also derived previously

P(B) = P(cancer) = P({BH, BL}) = P({BH}) + P({BL}) = 0.001519 + 0.001581 = 0.0031

So based on this it becomes quite easy to answer our new conditional question:

P(H | B) = P({BH}) / P(B) = 0.001581 / 0.0031 = 0.51

So the sensitivity of the test is 0.51 or 51%, which means that the test correctly detects about half of the people who have cancer.

Now let’s look back at P(B | H) and what do we see there

P(B | H) = P({BH}) / P(H) = 0.001581 / 0.091302 = 0.017316

The numerators are the same, which let’s us come up with a simple formula by forming the ratio between the two conditionals:

P(B | H) / P(H | B) = P({BH}) / P(H) * P(B) / P({BH}) = P(B) / P(H)

we can rewrite that as

P(B | H) = P(B)/P(H) * P(H | B)

So we now see that the answer to the crucial question – how likely is it that the patient has cancer conditional on a positive test – doesn’t depend just on the sensitivity of the test P(H | B) but also on the the ratio between the unconditional probabilities of having cancer P(B) and of receiving a high PSA signal P(H).

Readers with a background in statistics will recognize the above as the formula for Bayes’ Theorem. Next Wednesday we will go into more detail about what it means, but between now and then you should ask yourself how much the sensitivity of a test really tells you by itself.

Today’s Uncertainty Wednesday is a further continuation of the PSA Test example, but I got tired of the boring title and wanted to give the post as more, well, specific one. Last Wednesday I wrote that sensitivity and specificity are “widely used to assess the quality of medical tests.”

Here are their definitions taken from Wikipedia

Sensitivity = probability of a positive test given that the patient has the disease
Specificity = probability of a negative test given that the patient is well

Using our notation we can rewrite this as

Sensitivity = P(H | B), i.e. probability of a high PSA level signal (H) conditional on the patient having cancer (B)

Specificity = P(L | A), i.e. probability of a low PSA level signal (L) conditional on the patient being healthy (A)

We see that sensitivity is in a way the inverse of the question we were asking originally, which is P(B | H), i.e the probability of having cancer (B) conditional on a receiving a high PSA level signal (H).

So the question is what is the relationship between P(H | B) and P(B | H)?

To figure this out, let’s first see if we can derive P(H | B) from the elementary probabilities given originally, which I am repeating here

P({AL}) = P(healthy *and* low PSA) = 0.907179
P({AH}) = P(healthy *and* high PSA) = 0.089721
P({BL}) = P(cancer *and* low PSA) = 0.001519
P({BH}) = P(cancer *and* high PSA) = 0.001581

So let’t start by recalling out how likely it is that someone has cancer (B) which we also derived previously

P(B) = P(cancer) = P({BH, BL}) = P({BH}) + P({BL}) = 0.001519 + 0.001581 = 0.0031

So based on this it becomes quite easy to answer our new conditional question:

P(H | B) = P({BH}) / P(B) = 0.001581 / 0.0031 = 0.51

So the sensitivity of the test is 0.51 or 51%, which means that the test correctly detects about half of the people who have cancer.

Now let’s look back at P(B | H) and what do we see there

P(B | H) = P({BH}) / P(H) = 0.001581 / 0.091302 = 0.017316

The numerators are the same, which let’s us come up with a simple formula by forming the ratio between the two conditionals:

P(B | H) / P(H | B) = P({BH}) / P(H) * P(B) / P({BH}) = P(B) / P(H)

we can rewrite that as

P(B | H) = P(B)/P(H) * P(H | B)

So we now see that the answer to the crucial question – how likely is it that the patient has cancer conditional on a positive test – doesn’t depend just on the sensitivity of the test P(H | B) but also on the the ratio between the unconditional probabilities of having cancer P(B) and of receiving a high PSA signal P(H).

Readers with a background in statistics will recognize the above as the formula for Bayes’ Theorem. Next Wednesday we will go into more detail about what it means, but between now and then you should ask yourself how much the sensitivity of a test really tells you by itself.