Risk Function for a hypothesis test?

Question

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1. What the problem is asking

We have i.i.d. normal observations

[ X_1,\dots ,X_n \stackrel{\text{i.i.d.}}{\sim} N(\mu,\sigma^{2}),\qquad \sigma^{2}\ \text{known}. ]

We want to test

[ H_{0}:\mu =0\qquad\text{vs.}\qquad H_{1}:\mu\neq 0 ]

with the (two‑sided) decision rule

[ \mathcal R={\;|\bar X|\ge k\;}\qquad\text{(reject }H_{0}\text{)} . ]

The loss (cost) associated with a wrong decision is

(L(H_{0},\text{reject }H_{0}) = c_{10}) (type‑I error)
(L(H_{1},\text{retain }H_{0}) = c_{01}) (type‑II error)

All other outcomes have loss (0).

Under the alternative we do not know the true (\mu); instead we are told that

[ \mu\mid H_{1}\;\sim\;N(0,\tau^{2}) . ]

The task is to compute the risk function, i.e. the expected loss,

when the true state is (H_{0}) (i.e. (\mu =0)), and
when the true state is (H_{1}) (i.e. (\mu\neq 0) with the prior above).

2. Distribution of the test statistic

The sample mean is

[ \bar X = \frac{1}{n}\sum_{i=1}^{n}X_i . ]

Conditional on a fixed (\mu)

[ \bar X \mid \mu \;\sim\; N!\Bigl(\mu,\; \frac{\sigma^{2}}{n}\Bigr). ]

Marginalising over the prior (\mu\sim N(0,\tau^{2})) (only under (H_{1}))

Because a sum of independent normal variables is normal,

[ \bar X \mid H_{1}\;\sim\; N!\Bigl(0,\; \frac{\sigma^{2}}{n}+\tau^{2}\Bigr). ]

Both distributions will be needed.

3. Risk when the null hypothesis is true ((\mu =0))

When (\mu=0) the only way we incur loss is by rejecting (H_{0}).
Hence

[ R_{0}= \operatorname{E}\bigl[ L\mid \mu=0 \bigr] = c_{10}\; P\bigl(|\bar X|\ge k \mid \mu=0 \bigr). ]

Because (\bar X\mid \mu=0\sim N!\bigl(0,\sigma^{2}/n\bigr)),

[ \begin{aligned} P\bigl(|\bar X|\ge k \mid \mu=0\bigr) &= P!\left(\bar X\ge k\right)+P!\left(\bar X\le -k\right)
&= 2\Bigl[1-\Phi!\Bigl(\frac{k}{\sigma/\sqrt{n}}\Bigr)\Bigr]
&= 2\Bigl[1-\Phi!\Bigl(\frac{k\sqrt{n}}{\sigma}\Bigr)\Bigr], \end{aligned} ]

where (\Phi(\cdot)) is the standard normal cdf.

Therefore

[ \boxed{ \displaystyle R_{0}= c_{10}\; 2\Bigl[1-\Phi!\Bigl(\frac{k\sqrt{n}}{\sigma}\Bigr)\Bigr] } . ]

4. Risk when the alternative hypothesis is true

4.1 Conditional risk for a fixed non‑zero (\mu)

If the true mean were some specific (\mu\neq0), we would incur loss only when we retain (H_{0}) (i.e. (

\bar X

<k)). Thus

[ R_{1}(\mu)=c_{01}\;P\bigl(|\bar X|<k\mid \mu\bigr). ]

Using the conditional distribution (\bar X\mid \mu\sim N(\mu,\sigma^{2}/n)),

[ \begin{aligned} P\bigl(|\bar X|<k\mid \mu\bigr) &=P\bigl(-k<\bar X<k\mid \mu\bigr)
&= \Phi!\Bigl(\frac{k-\mu}{\sigma/\sqrt n}\Bigr) -\Phi!\Bigl(\frac{-k-\mu}{\sigma/\sqrt n}\Bigr). \end{aligned} ]

Hence

[ \boxed{\displaystyle R_{1}(\mu)=c_{01}\Bigl[ \Phi!\Bigl(\frac{k-\mu}{\sigma/\sqrt n}\Bigr) -\Phi!\Bigl(\frac{-k-\mu}{\sigma/\sqrt n}\Bigr) \Bigr] } . ]

4.2 Bayes risk – averaging over the prior (\mu\sim N(0,\tau^{2}))

The problem statement says “under (H_{1}) we also assume that (\mu\sim N(0,\tau^{2}))”.
Therefore the overall risk when the true state is (H_{1}) is the prior‑average of the conditional risk:

[ \begin{aligned} R_{1} &= \operatorname{E}{\mu}\bigl[ R{1}(\mu)\bigr]
&= c_{01}\; \operatorname{E}_{\mu}\bigl[ P(|\bar X|<k\mid\mu) \bigr] . \end{aligned} ]

Because the inner probability is just the marginal probability that the statistic falls in the acceptance region, we can replace the two‑step expectation by a single probability computed with the marginal distribution of (\bar X) under (H_{1}):

[ \bar X\mid H_{1}\;\sim\; N!\Bigl(0,\; \frac{\sigma^{2}}{n}+\tau^{2}\Bigr). ]

Thus

[ \begin{aligned} R_{1} &= c_{01}\; P\bigl(|\bar X|<k\mid H_{1}\bigr)\[4pt] &= c_{01}\Bigl[ \Phi!\Bigl(\frac{k}{\sqrt{\sigma^{2}/n+\tau^{2}}}\Bigr) -\Phi!\Bigl(\frac{-k}{\sqrt{\sigma^{2}/n+\tau^{2}}}\Bigr) \Bigr] \[4pt] &= c_{01}\Bigl[\,2\Phi!\Bigl(\frac{k}{\sqrt{\sigma^{2}/n+\tau^{2}}}\Bigr)-1\Bigr]. \end{aligned} ]

So the Bayes risk for the alternative is

[ \boxed{\displaystyle R_{1}=c_{01}\Bigl[\,2\Phi!\Bigl(\frac{k}{\sqrt{\sigma^{2}/n+\tau^{2}}}\Bigr)-1\Bigr] } . ]

5. Summary of the risk function

True state	Risk (expected loss)
(H_{0}) ((\mu =0))	(\displaystyle R_{0}=c_{10}\;2\bigl[1-\Phi!\bigl(\tfrac{k\sqrt n}{\sigma}\bigr)\bigr])
(H_{1}) ((\mu\neq0), with prior (\mu\sim N(0,\tau^{2})))	(\displaystyle R_{1}=c_{01}\Bigl[\,2\Phi!\bigl(\tfrac{k}{\sqrt{\sigma^{2}/n+\tau^{2}}}\bigr)-1\Bigr])

If one wishes the risk conditional on a specific non‑zero (\mu), replace the second line by

[ R_{1}(\mu)=c_{01}\Bigl[ \Phi!\Bigl(\frac{k-\mu}{\sigma/\sqrt n}\Bigr) -\Phi!\Bigl(\frac{-k-\mu}{\sigma/\sqrt n}\Bigr) \Bigr]. ]

6. Common Mistakes

Mistake	Why it is wrong	How to avoid it
Using the marginal variance (\sigma^{2}/n+\tau^{2}) for the null case	Under (H_{0}) the prior on (\mu) is not invoked; (\mu) is fixed at 0, so the variance is only (\sigma^{2}/n).	Remember the prior only applies under (H_{1}).
Omitting the “(-1)” term in the type‑II risk	The probability of not rejecting is (P(	\bar X	<k)=2\Phi(\cdot)-1), not just (2\Phi(\cdot)).	Write the acceptance probability as (\Phi(a)-\Phi(-a)=2\Phi(a)-1).
Treating (\bar X) as (N(0,\sigma^{2}/n+\tau^{2})) for every (\mu\neq0)	That distribution is the mixture (marginal) under the prior, not the conditional distribution given a particular (\mu).	Distinguish clearly: conditional on a fixed (\mu) → variance (\sigma^{2}/n); marginal (averaged over prior) → extra (\tau^{2}).
Confusing type‑I and type‑II losses	The loss (c_{10}) is paid only when we reject a true (H_{0}); (c_{01}) is paid only when we fail to reject a true (H_{1}).	Write the loss table explicitly before forming the risk.
Using a one‑sided critical region	The test is two‑sided, so the rejection region is (	\bar X	\ge k).	Ensure the probability calculations include both tails (multiply by 2 or use symmetry).

With the correct distributions and the proper handling of the prior, the risk expressions above follow directly.

Original question: Risk Function for a hypothesis test? on Cross Validated (Stats Stack Exchange), licensed CC BY-SA.